1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 7 InterviewSolution

301.	Calculate the pH of (i) 1 M `H_(2)SO_(4) " " (ii) 2 M H_(2)SO_(4) " " (iii) 10^(-2)M H_(2)SO_(4)` solutions. Given that the second ionization constant `(K_(a_(2))) " of" H_(2)SO_(4) "if" 10^(-2)`.
Answer» `H_(2)SO_(4)` ionizes in two steps as `H_(2)SO_(4) rarr H^(+)+HSO_(4)^(-) ` (almost completely ionized ) `HSO_(4)^(-) hArr H^(+) + SO_(4)^(2-)` (partially ionized) (i) In 1 M `H_(2)SO_(4), H^(+)` ions are obtained mainly from 1st ionization = 1 M and `H^(+)` ions obtained from 2nd step are negligible as compared to 1 M. Hence `[H^(+)] = 1 M , pH = - log 1 = 0 `. (ii) In 2 M `H_(2)SO_(4)` , again `H^(+)` ions are obtained mainly from 1st ionization as explained above. Hence, `[H^(+)]=2 M, pH = - log 2 = - 0.303`. Here note that there is nothing wrong with the negative value of pH but this offers no advantage over the actual value of `[H^(+)]`. (iii) In `10^(-2) M H_(2)SO_(4),H^(+)` ions obtained from 2nd ionization cannot be neglected in comparison to `10^(-2) M H^(+)` ions obtained from 1 st ionization. Hence, we have `{:(,H_(2)SO_(4),rarr,H^(+),+,HSO_(4)^(-),),("Initial",10^(-2)M,,,,,),("After 1 st ionization",~=0,,10^(-2)M,,10^(-2)M,),("Initial",HSO_(4)^(-),hArr,H^(+),+,SO_(4)^(-),),(,10^(-2)M,,,,,),("After ionization",10^(-2)-x,,10^(-2)+x,,x,):} (H^(+)=10^(-2) "from 1 st ionization + x from 2nd ionization")` `:. K_(a_(2))=([H^(+)][SO_(4)^(2-)])/([HSO_(4)^(-)])=((10^(-2)+x)(x))/(10^(-2)-x)=10^(-2)` (Given) `:. 10^(-2)x + x^(2)=10^(-4)-10^(-2)x or x^(2) + 2 xx 10^(-2)x-10^(-4) = 0 ` `:. x=(-b pmsqrt(b^(2)-4ac))/(2a) = (-2xx10^(-2)pmsqrt(4xx10^(-4)+4xx10^(-4)))/(2) = (-2xx10^(-2)+2.828xx10^(-2))/(2)` `=(0.828xx10^(-2))/(2)` (ignoring the -ve value ) `=0.414xx10^(-2) M` `:. [H^(+)]=10^(-2)+x=10^(-2)+0.414xx10^(-2) = 1.414xx10^(-2)` `:. pH = - log [H^(+)]=-log(1.414xx10^(-2))=1.85`

Discussion

302.	Classify each of the following substances into an acid or base or both and mention the concept/concepts on the basis of which you can do so. `(i) HCl(aq) " " (ii) NH_(3)(g) " " (iii) Na_(2)CO_(3)(aq) " " (iv)CH_(3)CO OH(aq)` `(v)CO_(2)(g) " " (vi)BF_(3) " " (vii)Ag^(+) " " (viii)CN^(-)` (ix)`H_(2)O " " (x)H_(2)SO_(4) " (xi)" HCO_(3)^(-) " (xii)"SiF_(4)`
Answer» (i) HCl (aq) - Acid (Arrhenius conecpt and Bronsted - Lowry concept) (ii) `NH_(3)(g) `-Base (Bronsted concept and Lewis concept) (iii) `Na_(2)CO_(3)(aq)`- Base (Bronsted concept) (iv) `CH_(3)CO OH(aq)`-Acid (Arrhenius concept and Bronsted concept) (v) `CO_(2)(g)`-Acid (Bronsted concept and Lewis concept) (vi) `BF_(3)`-Acid (Lewis concept) (vii) `Ag^(+)`- Acid (Lewis concept) (viii) `CN^(-)`- Base (Lewis concept) (ix) `H_(2)O`- Both acid and base, i.e., amphoteric (Bronsted concept) (x) `H_(2)SO_(4)`- Both acid and base, i.e., amphoteric (Bronsted concept) (xi) `CHO_(3)^(-)`- Both acid and base, i.e., amphoteric (Bronsted concept) (xii) `SiF_(4) ` - Acid (Lewis concept), as silicon can expand its octet .

Discussion

303.	Given a solution of acetic acid. How many times of the acid concentration, acetate salt should be added to obtain a solution with pH = 7.0 [`K_(a)` for dissociation of `CH_(3)CO OH=1.8xx10^(-5)]`
Answer» Suppose the concentration of the acetate salt added is x times the concentration of acetic acid solution to obtain `pH=7. "Then " [CH_(3)CO O^(-)]=x xx [CH_(3)CO OH] :. ([CH_(3)CO O^(-)])/([CH_(3)CO OH])=x` Now, for acid buffer, `pH = pK_(a) + log .(["Salt"])/("Acid"])` `:. 7 = - log (1.8xx10^(-5))+ log x or 7 = 4.76+ log x or log x = 2.24 or x = 1.73 xx 10^(2) = 173` times

Discussion

304.	Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionization constant of acetic acid is 1.75 x 10–5 . Also calculate the change in pH of the buffer if to 1 L of the buffer(i) 1 c.c. of 1 M NaOH are added(ii) 1 c.c. of 1 M HCl are added.Assume that the change in volume is negligible.(iii) What will be the buffer index of the above buffer?
Answer» Given, [CH₃COOH] = 0.1 M [CH₃COONa] = 0.15 M pK_a = 1.75 × 10^–15 For an acidic buffer pH = pK_a + log\(\frac{[Salt]}{[acid]}\) pK_a = – log [K_a] = – log [1.75 × 10^–15] = – log 1.75 + 5 log 10 = – 0.2430 + 5 = 4.757 From equation (i) pH = 4.757 + log\(\frac{(0.15)}{(1.10)}\) = 4.757 + 1761 = 4.933 (i) 1 cc of 1 M NaOH contains NaOH = 10^–3 mol. This will convert 10^–3mol of acetic acid into the salt so the salt formed = 10^–3 mol [Acid] = 0.10 – 0.001 = 0.099 M [Salt] = 0.15 + 0.001 = 0.151 M \(\therefore\) pH = pK_a + log\(\frac{[Salt]}{[acid]}\) = 4.757 + log\(\frac{0.151}{0.099}\) = 4.757 + 0.183 = 4.940 \(\therefore\) Increase in pH = 4.940 – 4.933 = 0.077 which is negligible (ii) 1 cc of 1 M HCl contains HCl = 10^–3 M This will convert 10^–3 mol CH₃COONa into CH₃COOH. \(\therefore\) [Acid] = 0.10 + 0.001 = 0.101 M [Salt] = 0.15 – 0.001 = 0.149 M pH = pK_a + log\(\frac{[salt]}{[acid]}\) = 4.757 + log\(\frac{0.149}{0.101}\) = 4.747 + 0.169 = 4.925 \(\therefore\) Decrease in pH = 4.933 – 4.925 = 0.008 which is neglibile (iii) No. of moles of HCl or NaOH added in buffer = 0.001 mol Change in pH ≃ 0.007 \(\therefore\) Buffer index = \(\frac{No.\,of\,moles\,of\,acidor\,base\,added}{change\,in\,pH}\) = \(\frac{0.001}{0.007}\) =\(\frac{1}{7}\) = 0.143

304.

Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionization constant of acetic acid is 1.75 x 10–5 . Also calculate the change in pH of the buffer if to 1 L of the buffer(i) 1 c.c. of 1 M NaOH are added(ii) 1 c.c. of 1 M HCl are added.Assume that the change in volume is negligible.(iii) What will be the buffer index of the above buffer?

Answer»

Given,

[CH₃COOH] = 0.1 M

[CH₃COONa] = 0.15 M

pK_a = 1.75 × 10^–15

For an acidic buffer

pH = pK_a + log\(\frac{[Salt]}{[acid]}\)

pK_a = – log [K_a]

= – log [1.75 × 10^–15]

= – log 1.75 + 5 log 10

= – 0.2430 + 5

= 4.757

From equation (i)

pH = 4.757 + log\(\frac{(0.15)}{(1.10)}\)

= 4.757 + 1761

= 4.933

(i) 1 cc of 1 M NaOH contains NaOH = 10^–3 mol. This will convert 10^–3mol of acetic acid into the salt so the salt formed = 10^–3 mol

[Acid] = 0.10 – 0.001 = 0.099 M

[Salt] = 0.15 + 0.001 = 0.151 M

\(\therefore\) pH = pK_a + log\(\frac{[Salt]}{[acid]}\)

= 4.757 + log\(\frac{0.151}{0.099}\)

= 4.757 + 0.183

= 4.940

\(\therefore\) Increase in pH = 4.940 – 4.933

= 0.077 which is negligible

(ii) 1 cc of 1 M HCl contains HCl = 10^–3 M

This will convert 10^–3 mol CH₃COONa into CH₃COOH.

\(\therefore\) [Acid] = 0.10 + 0.001 = 0.101 M

[Salt] = 0.15 – 0.001 = 0.149 M

pH = pK_a + log\(\frac{[salt]}{[acid]}\)

= 4.757 + log\(\frac{0.149}{0.101}\)

= 4.747 + 0.169

= 4.925

\(\therefore\) Decrease in pH = 4.933 – 4.925

= 0.008 which is neglibile

(iii) No. of moles of HCl or NaOH added in buffer = 0.001 mol

Change in pH ≃ 0.007

\(\therefore\) Buffer index = \(\frac{No.\,of\,moles\,of\,acidor\,base\,added}{change\,in\,pH}\)

= \(\frac{0.001}{0.007}\)

=\(\frac{1}{7}\) = 0.143

Discussion

305.	Calculate the concentration of `H_(3)O^(+)` ions in a mixture of 0.02 M acetic acid and 0.2 M sodium acetate. Given that the ionization constant `(K_(a))` for acetic acid is `1.8xx10^(-5)`.
Answer» `{:(,CH_(3)CO OH +H_(2)O,hArr,CH_(3)CO O^(-) +, H_(3)O^(+),),("Initial moles",0.02,,0,0,),("Moles at eqm.",0.02-x,,x,x,),(,CH_(3)CO ONa,rarr,0,0,),("Initial moles",0.2,,0,0,),("Moles at eqm.",0,,0.2,0.2,),(,,,,,):}` Thus, in the mixture solution, `[CH_(3)CO O^(-)]=0.2+x~=0.2 M (CH_(3)CO O^(-)" are obtained mainly from" CH_(3)CO ONa," therefore," x lt lt 0.2)` `[CH_(3)CO OH ] = 0.02 - x ~= 0.02 M` `K_(a)=([CH_(3)CO O^(-)][H_(3)O^(+)])/([CH_(3)CO OH]), i.e., 1.8xx10^(-5)=(0.2xx[H_(3)O^(+)])/(0.02)` or`[H_(3)O^(+)]=(1.8xx10^(-5)xx0.02)/(0.2) = 1.8xx10^(-6)M` Note. The given mixture is a buffer as discussed later in Art. 7.32 .

Discussion

306.	Calculate the percentage hydrolysis of decinormal solution of ammonium acetate given that `k_(a) = 1.75 xx 10^(-5), K_(b) =1.80 xx 10^(-5) and K_(w) = 1.0 xx 10^(-14)`
Answer» (i). Calculation of hydrolysis constant `(K_(h)).` Since ammonium acetate is salt of weak acid and weak base, `K_(h) =(K_(w))/(K_(a).K_(b)) = (1.0xx10^(-14))/(1.75xx10^(-5)xx1.80xx10^(-5))=3.175 xx10^(-5)` (II). Calculatation of degree of hydrolysis (h). `h=sqrt(K_(h))= (3.175 xx 10^(-5))^(1//2) =(31.75 xx 10^(-6))^(1//2) =5.63 xx 10^(-3).`

Discussion

307.	Predict if the solutions of the following salts are neutral, acidic or basic :NaCl, KBr, NaCN, NaOH H2SO4, NaNO2, NH4 NO3, KF
Answer» NaCl - Neutral KBr - Neutral NaCN - Basic NaOH - Basic H₂SO₄- Acidic NaNO₂- Basic NH₄NO₃ - Acidic KF - Basic

307.

Predict if the solutions of the following salts are neutral, acidic or basic :NaCl, KBr, NaCN, NaOH H2SO4, NaNO2, NH4 NO3, KF

Answer»

NaCl - Neutral

KBr - Neutral

NaCN - Basic

NaOH - Basic

H₂SO₄- Acidic

NaNO₂- Basic

NH₄NO₃ - Acidic

KF - Basic

Discussion

308.	13.5 g of an acid HA of molecular mass 135 was dissolved in 10 litres of aqueous solution Calculate the pH of the solution assuming the acid to be completely dissociated.
Answer» Correct Answer - `pH=2`

Discussion

309.	Find the pH of the following soluitons: (i) 3.2 g of hydrogen chloride dissolved in 1.0 L of water (ii) 0.28 g of potassium hydroxide dissolved in 1.0 L of water .
Answer» In both the problems , the volume of the solution is the same as the volume of water (i) Mass of hydrogen chloride =3.2 g Molar mass of hydrogen chloride = 36.5 g `mol^(-1)` No. of moles of hydrogen chloride `=((3.2 g))/((36.5 gmol^(-1)) = 0.0877 mol` Volume of solution = 1.0 L Molar concentration of solution`=((0.0877 mol))/((1.0L)) =0.0877 mol L^(-1) =0.0877 M` The acid is strong and is completely ionised in solution as : `HCloverset(aq)(to) underset(0.0877M)(H_(3)O^(+))+Cl^(-)(aq)` `[H_(3)O^(+)] = 0.0877 M` `pH=- log [H_(3)O^(+)] =- log (0.0877)=- log (8.77 xx 10^(-2))` `=(2-log 8.77) =(2-0.94) = 1.06` (ii) Mass of potassium hydroxide (KOH) = 0.28 g Molar mass of KOH = 56. 0 g `mol^(-1)` `:. [OH^(-)]` in the resulting solution `=0.1 xx 10^(-2) xx (1000)/(500) =2.0xx10^(-3) M` `[H_(3)O^(+)]=(K_(w))/[[OH^(-)]]=((10^(-14)M^(2)))/((2.0xx10^(-3)M))=5xx10^(-12)M` `pH=-log [H_(3)O^(+)] =- log (5.0 xx 10^(-12))` `=(12 - log 5) =(12-0.69897)= 11.30103`

Discussion

310.	The ionization constant of propanoic acid is `1.32xx10^(-5)`. Calculate the degree of ionization if its solution is 0.05 M. What will be its degree of ionization if the solution is 0.01 M in HCl also?
Answer» Assuming `alpha` to be very small, applying the formula directly, we have `alpha=sqrt(K_(a)//c)=sqrt((1.32xx10^(-5))//0.05)=1.62xx10^(-2)` `CH_(3)CH_(2)CO OH hArrCH_(3)CH_(2)CO O^(-)+ H^(+)` In presence of HCl, equilibrium will shift in the backward direction, i.e., concentration of `CH_(3)CH_(2)CO OH` will increase, i.e., amount dissociated will be less. If c is the initial concentration and x is the amount now dissociated, then at equilibrium `[CH_(3)CH_(2)CO OH]=c-x, [CH_(3)CH_(2)CO O^(-)]=x, [H^(+)]=0.01 + x ` `:. K_(a) = (x(0.01xx x))/(c-x)~=(x(0.01))/(c) or (x)/(c) =(K_(a))/(0.01)=(1.32xx10^(-5))/(10^(-2))=1.32xx10^(-3)`. But `(x)/(c) ` means `alpha`. Hence, `alpha = 1.32xx10^(-3)`.

Discussion

311.	What will be the resultant pH, when 200 mL of an aqueous solution of `HCI(pH=2.0)` is mixed with 300 mL of an aqueous solution of `NaOH(pH=12.0)` ?
Answer» pH of HCI solution `=2,[H_(3)O^(+)]` in solution `=10^(-2)M` Number of moles of `[H_(3)O^(+)]` in 200 mL of HCI solution `=(10^(-2) xx 200)/(1000) =0.2 xx 10^(-2) mol` pH of NaOH solution `=12 , [OH^(-)]` in solution `=10^(-2)M` Number of moles of `[OH^(-)]` in 300 mL of NaOH solution `=(10^(2)xx300)/(1000) =0.3 xx 10^(-2) mol.` Total volume of the solution after mixing `-(200 +300) =500 mL` Number of moles of NaOH left in the solution after neutralisation by the acid `=(0.3 xx 10^(-2) -0.2 xx 10^(-2)) =0.1 xx10^(-2) mol`

Discussion

312.	Calculate the degree of hydrolysis of 0.1 M solution of sodium acetate at `298 K : K_(a) = 1.8 xx 10^(-5)`.
Answer» (I). Calculation of hydrolysis constant `(K_(h)).` Since sodium acetate is a salt of strong base and weak acid `K_(h) =(K_(w))/(K_(a)) =(1.0xx10^(-14))/(1.8xx10^(-5)) =5.5 xx 10^(-10)` (II). Calculation of degree of hydrolysis (h). `K_(h) =5.5 xx 10^(-10) , C =0.1 M` Degree of hydrolysis (h)`=sqrt((K_(h))/(C))=sqrt((5.5xx10^(-10))/(0.1))=(55xx10^(-10))^(1//2) =7.42 xx10^(-5).`

Discussion

313.	Neutral solutions have pH = 7 at 298 K. A sample of pure water is found to have pH `lt ` 7. Does it mean that it is acidic ? Explain.
Answer» `pH lt 7` for pure `H_(2)O` shows that water is at a temperature higher than 298 K. It is neutral at all temperatures. At higher temperature, `H_(2)O` dissociates more to give larger concentrations of `H^(+)` ions and `OH^(-)` ions. Hence, `pH lt 7`. However, `[H^(+)]=[OH^(-)]` at all temperatures.

Discussion

314.	The pH of 0.004 M hydrazine solution is 9.7. Calculate its ionization constant Kb and pKb.
Answer» Given, pH = 9.7 \(\therefore\) pH = -log₁₀[H⁺] or [H⁺] = antilog (-pH) = antilog (-9.7) = 1.67 x 10^-10 \(\therefore\)[H⁺][OH^-] = Kw or [OH^-] = \(\frac{K_w}{(H^+)}\) = \(\frac{1\times10^{-14}}{1.67\times10^{-10}}\) = 5.98 x 10^-5 The concentration of the corresponding hydrazinium ion is also the same as that of OH⁻ ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004 M. \(\therefore\) For the reaction, NH₂NH₂ + H₂O ⇌ NH₂NH₄^⊕+ OH^⊕ K_b = \(\frac{[NH_4^+][OH^-]}{[NH_3]}\) = \(\frac{Ca.Ca}{C(1-a)}\) = 8.96 × 10⁻⁷ pK_b = -logK_b = -log(8.96 x 10^-7) = log 8.96 + 7 log 10 = - 0.9523 + 7 = 6.047

314.

The pH of 0.004 M hydrazine solution is 9.7. Calculate its ionization constant Kb and pKb.

Answer»

Given, pH = 9.7

\(\therefore\) pH = -log₁₀[H⁺]

or [H⁺] = antilog (-pH)

= antilog (-9.7)

= 1.67 x 10^-10

\(\therefore\)[H⁺][OH^-] = Kw

or [OH^-] = \(\frac{K_w}{(H^+)}\)

= \(\frac{1\times10^{-14}}{1.67\times10^{-10}}\)

= 5.98 x 10^-5

The concentration of the corresponding hydrazinium ion is also the same as that of OH⁻ ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004 M.

\(\therefore\) For the reaction,

NH₂NH₂ + H₂O ⇌ NH₂NH₄^⊕+ OH^⊕

K_b = \(\frac{[NH_4^+][OH^-]}{[NH_3]}\)

= \(\frac{Ca.Ca}{C(1-a)}\)

= 8.96 × 10⁻⁷

pK_b = -logK_b

= -log(8.96 x 10^-7)

= log 8.96 + 7 log 10

= - 0.9523 + 7

= 6.047

Discussion

315.	Find the pH of the following solution:(a) 3.2 g of hydrogen chloride dissolved in 1.00 L of water.(b) 0.28 g of potassium hydroxide dissolved in 1.00 water.
Answer» (a) Molarity of HCl = \(\frac{3.2}{36.5}\) x 1 0.089 M = 8.9 x 10^-2 Since HCl is completely ionized, [H⁺] = 8.9 x 10^-2 mol L^-1 pH = - [log H⁺] = -[log 8.9 x 10^-2] = -[-2 + 0.7762] = 1.2238 (b) Molarity of KOH = \(\frac{0.28}{56\times1}\) = \(\frac{1}{200}\) = \(\frac{1}{2\times10^2}\) = 0.5 x 10^-2 = 5 x 10^-2 mol l^-1 [H⁺] = \(\frac{K_w}{[OH^-]}\) [H⁺] = \(\frac{1.0\times10^{-14}}{5\times10^{-2}}\) = 0.2 x 10^-11 = 2 × 10⁻¹² pH = − log (2 × 10⁻¹²) = −(−12 + 0 ∙ 3010) = 11.699

315.

Find the pH of the following solution:(a) 3.2 g of hydrogen chloride dissolved in 1.00 L of water.(b) 0.28 g of potassium hydroxide dissolved in 1.00 water.

Answer»

(a) Molarity of HCl

= \(\frac{3.2}{36.5}\) x 1

0.089 M = 8.9 x 10^-2

Since HCl is completely ionized,

[H⁺] = 8.9 x 10^-2 mol L^-1

pH = - [log H⁺]

= -[log 8.9 x 10^-2]

= -[-2 + 0.7762] = 1.2238

(b) Molarity of KOH

= \(\frac{0.28}{56\times1}\) = \(\frac{1}{200}\) = \(\frac{1}{2\times10^2}\)

= 0.5 x 10^-2

= 5 x 10^-2 mol l^-1

[H⁺] = \(\frac{K_w}{[OH^-]}\)

[H⁺] = \(\frac{1.0\times10^{-14}}{5\times10^{-2}}\)

= 0.2 x 10^-11

= 2 × 10⁻¹²

pH = − log (2 × 10⁻¹²)

= −(−12 + 0 ∙ 3010)

= 11.699

Discussion

316.	The ionization constant of propionic acid is `1.32xx10^(-5)`. Calculate the degree of ionization of the acid in its `0.05`M solution and also its pH. What will be its degree of ionization in the solution of `0.01N HCI` ?
Answer» (I). Caculation of `alpha` for propionic acid According to Ostwald Dilution Law, `alpha +((K_(a))/(C))^(1/2) = ((1.32 xx 10^(-5))/(0.05))^(1/2) =(2.64 xx 10^(-4))^(1//2) = 1.62 xx 10^(-2)` (II). Calculation of pH of the solution `[H^(+)] =(K_(a) xx C)^(1//2) =(1.32 xx 10^(-5)xx 5xx 10^(-2))^(1//2)` `=(6.6 xx 10^(-7))^(1//2) = (66 xx 10^(-8))^(1//2) =8.124 xx 10^(-4)` `pH =- log (8.124 xx 10^(-4)) =- (log 8.124 -4 log 10)` `=(4- log 8.124) = (4-0.909) = 3.09` (III). Calculation of `alpha` for propionic acid in 0.01 M HCI solution. `CH_(3)CH_(2)COOH hArr CH_(3)CH_(2)COO^(-) + H^(+)` In the pressure of HCI. the ionisation of `CH_(3)CH_(2)COOH` will decrease. If C is the initial concentration of acid adn x is the amount dissociated at equilibrium `[CH_(3)CH_(2)COOH] =C- x , [CH_(3)CH_(2)COO^(-)] = x , [H^(+)] = 0.01 +x` `K_(a) =[[CH_(3) CH_(2)COO^(-)][H^(+)]]/[[CH_(3)CH_(2)COOH]] =((x)xx(0.01 +x))/((C -X)) = (x(0.01))/(C)` `" or "" " (x)/(C) =(K_(a))/(0.01) =(1.32 xx 10^(-5))/(10^5) =(1.32 xx 10^(-5))/(10^(-2)) =1.32 xx 10^(-3) , alpha =(x)/(C) =1.32 xx 10^(-3)`

Discussion

317.	Solution of `0.1 N NH_(4)OH` and `0.1 N NH_(4)Cl` has `pH 9.25`, then find out `K_(b)` of `NH_(4)OH`.A. `9.25`B. `4.75`C. `3.75`D. `8.25`
Answer» Correct Answer - B It is a basic buffer. This, `pOH=pK_(b)+log.(C_("salt"))/(C_("base"))` For any aqueous solution at 298 K, we have `pH+pOH=pK_(w)=14` or `pOH=14-pH` `= 14-9.25` `=4.75` we are given `C_("salt")=C_("base")`. Thus, `log.(C_("salt"))/(C_("base"))=log 1= 0` Hence, `pOH= pK_(b)=4.75`.

Discussion

318.	The concentration of water molecules in pure water at 298 K isA. `10^(-7)M`B. `55.5M`C. `5.55M`D. `7.26M`
Answer» Correct Answer - B Molarity is defined as the number of moles of solute per liter of solution, Thus, the molarity of pure water means the number of moles of water in one litre of water. Volume `(H_(2)O)=1 L=1000 ml` Density `(H_(2)O)= 1 g mL^(-1)` Mass `(H_(2)O)=` Density x Volume `= (1g mL^(-1))(1000 ml)` `= 1000 g` Number of moles of `H_(2)O = ("Mass")/("Molar mass")` `= (1000g)/(18g mol^(-1))=55.5 mol` Since we have `55.5` mol of `H_(2)O` in 1 L of water, the molarity of pure water is `55.5` mol of `H_(2)O` in 1 L of water, the molarity of water is `55.5 mol L^(-1)`.

Discussion

319.

Determine the degree of ionization and pH of a 0.05 M of ammonia solution. The ionization constant of ammonia is 1.77 x 10-5 . Also calculate the ionization constant of conjugate acid of ammonia.

Answer»

Given, K_b = 1.77 × 10⁻⁵

C = 0.05 M

The ionization of NH₃in water is represented by the equation:

NH₃ + H₂O ⇌ NH₄⁺ + OH^-

Initial	C	0	0
Conc.
Equilibrium	C(1 - α)	Cα	Cα
Conc.

Where α = Degree of ionization

C = concentration of solution

Ionization constant of ammonia

K_b = \(\frac{[NH_4^+][OH^-]}{[NH_3]}\)

= \(\frac{Ca.Ca}{C(1-a)}\)

= \(\frac{Ca^2}{(1-a)}\)

The value of α is small, so, the equation can be simplified by neglecting α in comparison to 1 in the denominator on right hand side of the equation.

\(\therefore\) k_b = Cα²

or α = \(\sqrt{\frac{K_b}{C}}\)

= \(\sqrt{\frac{1.77\times10^{-5}}{0.05}}\)

= 0.018

[OH^-] = Cα

0.05 × 0.018

= 9.4 × 10⁻⁴M

\(\therefore\) [H⁺][OH^-] = K_w

or [H⁺] = \(\frac{K_w}{[OH^-]}\) = \(\frac{10^{-14}}{9.4\times10^{-4}}\)

1.06 × 10⁻¹¹

Now, pH = −log₁₀ [H⁺]

= -log(1.06 x 10^-11)

= -log1.06 + 11 log10

= - 0.0253 + 11

= 10.9747

Now, using the relation for conjugate acid – base pair,

K_a x K_b = K_w

K_a =\(\frac{K_w}{K_b}\)

= \(\frac{10^{-4}}{1.77\times10^{-5}}\)

= 5.64 x 10^-10

Discussion

320.	What is the hydrogen ion concentration of a solution with pH value 5.6 ?
Answer» Correct Answer - `2.5 xx 10^(-6) M` pH of the solution =5.6 or - log `[H^(+)] =5.6` `log[(1)/(H^(+))] =5.6 " or " [(1)/(H^(+))] = " Antilog " 5.6 =398107.2` `[H^(+)] =(1)/(398107. 2) =2.5 xx 10^(-6) M`

Discussion

321.	Determine the degree of ionisation and pH of a 0.05 M ammonia solution. Also calculate ionisation constant of the conjugate acid ammonia. Given that `K_(b)` for `NH_(3)" is "1.77 xx 10^(-5)`.
Answer» Calculation of degree of ionisation `(alpha)`. The ionisation of `NH_(3)` may be represented as `NH_(3) +H_(2)OhArr NH_(4)^(+) +OH^(-)` The dissociation constant for the base may be given as : `K_(b) =(Calpha^(2))/(1-alpha)=Calpha^(2)` or " "`alpha = ((K_(b))/(C))^(1//2) =((1.77xx10^(-5))/(0.05))^(1//2)=0.018` (II) Calculation of pH of solution `[OH^(-)] =Calpha = 0.05 xx0.018 =9.4 xx 10^(-4) M` `[H^(+)]=(K_(w))/[[OH^(-)]] = (10^(-14))/(9.4 xx10^(-4)) =1.06 xx 10^(-11)` `pH =- log [H^(+)] =- log (1.06 xx 10^(-11)) = 10.97` (III). Caculation of ionisation constant of conjugate acid `NH_(4)^(+)` ions We know that for conjugate base-acid `K_(a) xxK_(b) =K_(w) or K_(a)=(K_(w))/(K_(b)) , K_(a) =(10^(-14))/(1.77xx10^(-5)) =5.64 xx 10^(-10)`

Discussion

322.	For the reaction `CO(g) +H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)` at a given temperature the equilibrium amount of `CO_(2)` (g) can be increased byA. adding a suitable catalystB. adding an inert gasC. decreasing the volume of the containerD. increasing the amount of CO (g)
Answer» Correct Answer - D Catalyst has no effect on equilibrium . Decreasing the volume or pressure has no effect because `n_(r)=n_(p)`. Therefoer , the equilibrium shifts in the forward direction by increasing the conc. Of reatant CO (g).

Discussion

323.	Which of the following statement (s) is (are) correct ? (1) . The pH of `1.0 xx 10^(-8)` M solution of HCI is 8 (2). The conjugate base of `H_(2)PO_(4)^(-) " is " HPO_(4)^(2-)` (3). Autoprotolysis constant of water increases with temperature (4). When a solution of a weak monoprotic acid is titrated against a strong base at half neutralization point pH`= (1//2) pK_(a)`A. `2,3`B. `1,2,3`C. `3,4`D. `2,3,4`
Answer» Correct Answer - A Ph of 10^(-8)` M HCI is not 8 bu it is less than 7. Only choices 2,3 are correct.

Discussion

324.	The `pH` of `0.05M` aqueous solution of diethy`1` amine is `12.0` . Caluclate `K_(b)`.
Answer» pH of solution =12.0 (given) , `"But"" "pH +pOH =14 :. pOH =14 -pH =14 -12 =2` `" or "" "[OH^(-) (aq)] =10^(-2) M` the dissociation of `(C_(2)H_(5))_(2)` NH is represented as : `{:(,(C_(2)H_(5))_(2)NH+H_(2)O,hArr,(Cr_(2)H_(5))_(2)NH_(2)^(+),+,OH^(-)),("Initial no. of moles",0.05M,,0,,0),("No. of moles at the",(0.05M-0.01),,0.01M,,0.01M),("equilibrium point",=0.04M,,,,):}` Applying Law of chemical equilibrium , `K_(b) =[[(C_(2)H_(5))_(2)NH_(2)^(+)][OH^(-)]]/[[(C_(2)H_(5))_(2)NH]] =((0.01M) xx (0.01M))/((0.04 M)) =2.5 xx 10^(-3)M`

Discussion

325.	Write the conjugate acid for the following (a) Cl–, (b) NO3–, (c) HSO4–, (d) HCO3–, (e) SO42-, (f) CO32-, (g) NH2-, (h) N3- , (i) O2-, (j) H2O, (k) NH3.
Answer» Conjugate acid (a) Cl^–→ HCl (b) NO₃^–→ HNO₃ (c) HSO₄^–→ H₂SO₄ (d) HCO₃^– → H₂CO₃ (e) SO₄^2- → HSO₄^- (f) CO₃^2- → HCO₃^- (g) NH^2- → NH₂^- (h) N^3- → HN₂^- (i) O^2- → OH^- (j) H₂O → H₃O⁺ (k) NH₃ → NH₄⁺

Discussion

326.	What is the conjugate acid and conjugate base of HCO3 - ion ?
Answer» Conjugate acid is H₂CO₃ a conjugate acid is CO₃ ^2-

Discussion

327.	Write the conjugate base of HCO3
Answer» CO^2-₃ is conjugate base of HCO₃

Discussion

328.	Calculate the pH of 0.001 NHCl.
Answer» pH = – log₁₀ [H⁺] = – log₁₀ [10^-3] = 3log₁₀ = 3 × 1 = 3

Discussion

329.	The pH of a solution on addition of 50 ml. water slowly to a solution(a) 1(b) 5(c) 7(d) 10
Answer» The answer is (c) 7

Discussion

330.	At `450 K, K_(p)=2.0xx10^(10)//` bar for the given reaction at equilibrium. `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)` What is `K_(c )` at this temperature?
Answer» Correct Answer - `7.47 xx 10^(11) M^(-1)` For the given reaction, `Deltan = 2-3 = -1` `T = 450 K` `R = 0.0831 "bar" L "bar" K^(-1) "mol"^(-1)` `K_(P) = 2.0 xx 10^(10) "bar"^(-1)` We know that, `K_(P) = K_(c) (RT) Deltan` `rArr 2.0 xx 10^(10) "bar"^(-1) = K_(c)(0.0831 L "bar"K^(-1) "mol"^(-1) xx 450 K)^(-1)` `rArr K_(c) = (2.0 xx 10^(10) "bar"^(-1))/((0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)^(-1))` `= 2.0 xx 10^(10) "bar"^(-1) = K_(c) (0.0831L "bar" K^(-1) "mol"^(-1) xx 450K)` `rArr K_(c) = (2.0 xx 10^(10)"bar"^(-1))/((0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)^(-1))` `= (2.0 xx 10^(10) "bar"^(-1))(0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)` `= 74.79 xx 10^(10) L "mol"^(-1)` `= 7.48 xx 10^(11) L "mol"^(-1)` `= 7.48 xx 10^(11) M^(-1)`

Discussion

331.	The reaction `2SO_(2)+2O_(2)hArr 2SO_(3)` will be favoured byA. high temperature and low pressureB. low temperature and high pressureC. high temperature and high pressureD. low temperature and low pressure.
Answer» Correct Answer - B Since reaction is exothermic it is favoured at low temperature. There is a decrease in number of moles hence reaction is favoured at high pressure.

Discussion

332.	What is chemical equilibrium ?
Answer» A dynamic state of a chemical reaction at which the rate of the forward reaction is equal to the rate of backward reaction is called chemical equilibrium.

Discussion

333.	Write any two factors which affect ionization.
Answer» (i) Concentration : In a solution, on increasing the concentration of electrolyte, ionization decreases. The extent of ionization of an electrolyte is inversely proportional to the concentration of its solution. (ii) Temperature : The degree of ionization increases with increase in temperature.

333.

Write any two factors which affect ionization.

Answer»

(i) Concentration : In a solution, on increasing the concentration of electrolyte, ionization decreases. The extent of ionization of an electrolyte is inversely proportional to the concentration of its solution.

(ii) Temperature : The degree of ionization increases with increase in temperature.

Discussion

334.	What are reversible and irreversible reactions?
Answer» Reversible reaction: A reaction which takes place in both forward and backward direction is called reversible reaction. Irreversible reaction: A reaction in which entire amount of reactant is changed into product and no reaction from product side occurs in called irreversible reaction.

334.

What are reversible and irreversible reactions?

Answer»

Reversible reaction: A reaction which takes place in both forward and backward direction is called reversible reaction.

Irreversible reaction: A reaction in which entire amount of reactant is changed into product and no reaction from product side occurs in called irreversible reaction.

Discussion

335.	What is the effect of adding an inert gas a) at constant volume b) at constant pressure on the following equation: PCl5 (g) PCl3(g) +Cl2(g)
Answer» a) At constant vol. the total pressure would increase due to inert gas addition but the conc. of the reactants and products will do not change hence no effect on equilibrium b) At const. pressure the vol. of the system would increase. This results in decrease in no. of the moles of reactants per unit vol. Hence, reaction will shift in forward direction.

335.

What is the effect of adding an inert gas a) at constant volume b) at constant pressure on the following equation: PCl5 (g) PCl3(g) +Cl2(g)

Answer»

a) At constant vol. the total pressure would increase due to inert gas addition but the conc. of the reactants and products will do not change hence no effect on equilibrium

b) At const. pressure the vol. of the system would increase. This results in decrease in no. of the moles of reactants per unit vol. Hence, reaction will shift in forward direction.

Discussion

336.	Which of the following reactions will get affected by increasing the pressure ? Also, mention whether change will cause the reaction to go into forward or backward direction.(i) COCl2 ➝ CO(g) + Cl2(g)(ii) CH4 (g) + 2S2 (g) ➝ CS2 (g) + 2H2S(g)(iii) CO2(g) + C(s) ➝ 2CO(g)(iv) 2H2 (g) + CO(g) ➝ CH3OH(g)(v) CaCO3 (s) ➝ CaO(s) + CO2 (g)(vi) 4NH3 (g) + SO2 (g) ➝ 4NO(g) + 6H2O(g)
Answer» (i) COCl₂ ➝ CO(g) + Cl₂(g) If pressure increases, the reaction will go into backward direction. (ii) CH₄ (g) + 2S₂ (g) ➝ CS₂ (g) + 2H₂S(g) If pressure increases, the reaction will go into backward direction. (iii) CO₂(g) + C(s) ➝ 2CO(g) If the pressure increases, the reaction will go into backward direction. (iv) 2H₂ (g) + CO(g) ➝ CH₃OH(g) If pressure increases, the reaction will go in to forward direction. (v) CaCO₃ (s) ➝ CaO(s) + CO₂ (g) If pressure increases, the reaction will go in to backward direction. (vi) 4NH₃ (g) + SO₂ (g) ➝ 4NO(g) + 6H₂O (g) If pressure increases, the reaction will go into backward direction.

Discussion

337.	which of the following reactions will get affected by increasing pressure? Mention whether the change will cause the reaction to go into forward or backward direction.(i) COCl2 (g) ----> CO (g) + Cl2(g) (ii) CO2 (g) +C(s) -----> 2CO (g) (iii) 2H2 (g) +CO (g) -------> CH3OH (l)
Answer» (i) np >nr Direction will be backward. (ii) np >nr Direction will be backward. (iii) np

Discussion

338.	Which of the following reaction will not affected on increasing the pressure ?A. `2H_(2(g))+CO_((g))hArrCH_(3)OH_((g))`B. `4NH_(3(g))+5O_(2(g))hArr4NO_((g))+6H_(2)O_((g))`C. `CH_(4(g))+2S_(2(g))hArrCS_(2(g))+2H_(2)S_((s))`D. `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`
Answer» Correct Answer - C Since number of moles of gaseous reactants and products are same, the reaction will not be affected by changing the pressure.

Discussion

339.	Ionisation constant of `CH_(3)CO O H ` is `1.7xx10^(-5)` and concentration of `H^(+)` ions is `3.4 xx 10^(-4)`. Then initial concentration of `CH_(3) C O O H` molecules isA. `3.4xx10^(-4)`B. `3.4xx10^(-3)`C. `6.8xx10^(-3)`D. `1.7xx10^(-3)`
Answer» Correct Answer - C `{:(,XH_(3)CO OH" " hArr" " CH_(3)CO O^(-) + H^(+)),("Initial conc."," "a),("At. eqm.",(a-3.4 xx 10^(-4)) " "3.4xx10^(-4)" " 3.4 xx 10^(-4)):}` `((3.4xx10^(-4))(3.4xx10^(-4)))/((a-3.4 xx 10^(-4))) = 1.7 xx 10^(-5) ` (Given) or, `(a-3.4xx10^(4))=((3.4xx10^(-4))^(2))/(1.7xx10^(-5))=6.8xx10^(-3)` or `a=6.8xx10^(-3)+3.4xx10^(-4)~~ 6.8 xx 10^(-3)`

Discussion

340.	In the following reactions, ZnO is respectively acting as a/an `(i) ZnO + Na_(2)O rarr Na_(2) ZnO_(2)` `(ii) NzO + CO_(2) rarr ZnCO_(3)`A. Acid and acidB. Acid and baseC. Base and acidD. Base and base
Answer» Correct Answer - B In (i) , NzO acts as an acidic oxide because `Na_(2)O` is a basic oxide In (ii), ZnO acts as a basic oxide because `CO_(2)` is an acidic oxide Thus, ZnO is an amphoteric oxide.

Discussion

341.	Solubilirty product constants `(K_(sp))` of salts of types `MX, MX_(2)`, and `M_(3)X` at temperature T are `4.0xx10^(-8), 3.2xx10^(-14)`, and salts of temperature T is in the orderA. `MX_(2)gtM_(3)XgtMX`B. `M_(3)XgtMX_(2)gtMX`C. `MXgtM_(3)XgtMX_(2)`D. `MXgtMX_(2)gtM_(3)X`
Answer» Correct Answer - C Let us assume that S is the solubility. `MX(s)hArr M^(+)underset(S)((aq.))+X^(-)underset(S)((aq.))` `K_(sp)=C_(M^(+))C_(X^(-))=(S)(S)` or `S=sqrt(K_(sp))=sqrt(4.0xx10^(-8))` `=2xx10^(-4)M` `MX_(2)(s)hArrM^(2+)underset(S)((aq.))+2X^(-)underset(2S)((aq.))` `K_(sp)=C_(M^(2+))C_(X^(-))^(2)=(S)(2S)^(2)=4S^(3)` or `S=((K_(sp))/(4))^(1//3)= ((3.2xx10^(-4))/(4))^(1//3)` `=2xx10^(-5)M` `M_(3)X(s)hArr underset(3S)(3M^(+)(aq.))+X^(3-)underset(S)((aq.))` `K_(sp)=C_(M^(+))^(3)C_(X^(3-))` `=(3S)^(3)(S)=27S^(4)` or `S=((K_(sp))/(27))^(1//4) = ((2.7xx10^(-15))/(27))^(1//4)` `=10^(-4)M` Thus, solubility order is `MXgt M_(3)XgtMX_(2)`

Discussion

342.	Solubility product constants `(K_(sp))` of salts of types MX, `MX_(2)` at `M_(3)X` at temperature T are `4.0xx10^(-8),3.2xx10^(-14) and 2.7xx10^(-15)` respectively. Solubility (mol `dm^(-3)`) of the salts at temperature T are in the orderA. `MX gt MX_(2) gt M_(3)X`B. `M_(3)X gt MX_(2) gt MX`C. `MX_(2) gt M_(3)X gt MX`D. `MX gt M_(3)X gt MX_(2)`
Answer» Correct Answer - D `MX hArr underset(s) M^(+)+underset(x)X^(-), K_(sp)=s^(2)` or `s=sqrt(K_(sp))=sqrt(4xx10^(-8))=2xx10^(-4)M` `MX_(2)hArr underset(s)overset(2+)M + 2 underset(2s)X^(-), K_(sp)=s(2s)^(2)=4s^(3)` `:. S=((K_(sp))/(4))^(1//3)=((3.2xx10^(-14))/(4))^(1//3)` `=2xx10^(-5)M` `M_(3)X hArr 3 underset(3s)M^(+)+ underset(s)X^(3-)`, `K_(sp)=(3s)^(3)(s)=27s^(4)` `:. s=((K_(sp))/(27))^(1//4)=((2.7xx10^(-15))/(27))^(1//4)=10^(-4) M` Thus, the solubilities are in the order `2xx10^(-4) gt 10^(-4) gt 2 xx 10^(-5)`, i.e., `MX gt M_(3)X gt MX_(2)`.

Discussion

343.	The solubility product constant `(K_(sp))` of salts of types `MX, MX_(2)`, and `M_(3)X` at temperature `T` are `4.0 xx 10^(-8), 3.2 xx 10^(-14)`, and `2.7 xx 10^(-15)`, respectively. The solubilities of the salts at temperature `T` are in the orderA. `MX gt MX_(2) gt M_(3)X`B. `M_(3)X gt MX_(2) gt MX`C. `MX_(2) gt M_(2)X gt MX`D. `MX gt M_(3)X gt MX_(2)`
Answer» Correct Answer - D `MX hArr underset((S_(1)))(M^(+)) +underset((S_(2)))(X^(-))` `S_(1) =(K_(sp))^(1//2) =(4 xx 10^(-8))^(1//2)` `=2xx 10^(-4).` `MX_(2) hArr underset((S_(2)))(M^(+)) +underset((2S_(2)))(2X^(-))` `5S_(2)^(3) =K_(sp) " or " S_(2) =((K_(sp))/(4))^(1/3)` `S_(2) =((3.2 xx 10^(-14))/(4))^(1/3) =2xx 10^(-5)` `M_(3)X hArr underset((3 S_(3)))(3M^(+)) + underset((S_(3)))(X^(3-))` `27S_(3)^(4) =K_(sp) " or " S_(3) =((K_(sp))/(27))^(1/4)` `S_(2) =((2.7 xx 10^(-5))/(27)) =10^(-4)` The correct order is : `MX gt M_(3)X gt MX_(2)`

Discussion

344.	In the following questions, two statements are given; one in assertion (A) column while the other in reason (R ) column. Examine the statements carefully and mark the correct answer according to the instructions given below: If both (A) and (R) are correct and (R) is the correct explanation of (A) If both (A) and (R) are correct but (R) is not the correct explanation of (A)If (A) is correct but (R) is wrongIf (A) is wrong but (R) is correct(A) When small amount of an acid or base is added to pure water its PH undergoes a change( R) Addition of an acid or a base increases the degree of ionization of water
Answer» If both (A) and (R) are correct and (R) is the correct explanation of (A)

Discussion

345.	A solution has `pH=5`, it is diluted `100` times, then it will becomeA. neutralB. basicC. unaffectesD. more acidic
Answer» Correct Answer - A `C_(H^(+))=10^(-pH)mol L^(-1)` Thus, pH = 5 means `C_(H^(+))= 10^(-5)mol L^(-1)` According to molarity equation, we have `M_(i)V_(i)= M_(f)V_(f)` `(10^(-5)mol L^(-1))(x)= M_(f)(100 x)` `:. M_(f)=10^(-7)mol L^(-1)` Under these conditions, the contribution of `H^(+)` from `H_(2)O` cannot be neglected. This will make the solutions slightly. acidic but practically neutral. In fact, its pH is `6.699~~7`.

Discussion

346.	`2mol` each of `A` and `B` are taken in a container to carry out the following reaction: `2A(g)+B(g)hArr2C(g)+2D(g)` When the system attains equilibrium, we haveA. `[A]lt[B]`B. `[A]gt[B]`C. `[A]=[B]`D. `[A]=[B]=[C]=[D]`
Answer» Correct Answer - A According to the stoichiometry of the reaction, for `1 mol` of `B` consumed, `2 mol` of `A` are converted into products. Thus, at equilibrium, the concentration of `A` will be less than that of B as we start with equal concentrations of `A` and `B`.

Discussion

347.	Which of the following anions is the weakest base ?A. `C_(2)H_(5)O^(-)`B. `CN^(-)`C. `CH_(3)COO^(-)`D. `NO_(3)^(-)`
Answer» Correct Answer - D The stronger the conjugate acid, the weaker the conjugate base. Since `HNO_(3)` is the strongest acid, its conjugate base `(NO_(3)^(-))` is the weakest base.

Discussion

348.	At 500 K, the equilibrium costant for the reaction `H_(2(g))+I_(2(g))hArr2HI_((g))" is "24.8" If "(1)/(2)mol//L` of HI is present at equilibrium, what are the concentrations of `H_(2)andI_(2)`, assuming that we started by taking HI and reached the equilibrium at 500 K ?A. `0.068"mol L"^(-1)`B. `1.020"mol L"^(-1)`C. `0.10"mol L"^(-1)`D. `1.20"mol L"^(-1)`
Answer» Correct Answer - C `{:(,2HI_((g)),hArr,H_(2(g)),+,I_(2(g)),,K_(c)=24.8),("Intial conc.",1,,0,,0,),("At equilibrium",0.5,,x,,x,):}` `K_(c)=([H_(2)][I_(2)])/([HI]^(2))or(1)/(24.8)=(x.x)/(0.5xx0.5)orx^(2)=(0.25)/(24.8)=0.010` `x=0.10"mol L"^(-1)`

Discussion

349.	How does common ion affect the solubility of electrolyte?
Answer» The solubility of electrolyte decreases due to common ion effect.

Discussion

350.	Equilivalent amounts of `H_(2)` and `I_(2)` are heated in a closed vessel till equilibrium is obtained. If `80%` of the hydrogen is converted to `HI`, the `K_(c)` at this temperature isA. 64B. 16C. `0.25`D. 14
Answer» Correct Answer - A

Discussion

Explore topic-wise InterviewSolutions in .

Calculate the pH of (i) 1 M `H_(2)SO_(4) " " (ii) 2 M H_(2)SO_(4) " " (iii) 10^(-2)M H_(2)SO_(4)` solutions. Given that the second ionization constant `(K_(a_(2))) " of" H_(2)SO_(4) "if" 10^(-2)`.

Given a solution of acetic acid. How many times of the acid concentration, acetate salt should be added to obtain a solution with pH = 7.0 [`K_(a)` for dissociation of `CH_(3)CO OH=1.8xx10^(-5)]`

Calculate the concentration of `H_(3)O^(+)` ions in a mixture of 0.02 M acetic acid and 0.2 M sodium acetate. Given that the ionization constant `(K_(a))` for acetic acid is `1.8xx10^(-5)`.

Calculate the percentage hydrolysis of decinormal solution of ammonium acetate given that `k_(a) = 1.75 xx 10^(-5), K_(b) =1.80 xx 10^(-5) and K_(w) = 1.0 xx 10^(-14)`

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13.5 g of an acid HA of molecular mass 135 was dissolved in 10 litres of aqueous solution Calculate the pH of the solution assuming the acid to be completely dissociated.

Find the pH of the following soluitons: (i) 3.2 g of hydrogen chloride dissolved in 1.0 L of water (ii) 0.28 g of potassium hydroxide dissolved in 1.0 L of water .

The ionization constant of propanoic acid is `1.32xx10^(-5)`. Calculate the degree of ionization if its solution is 0.05 M. What will be its degree of ionization if the solution is 0.01 M in HCl also?

What will be the resultant pH, when 200 mL of an aqueous solution of `HCI(pH=2.0)` is mixed with 300 mL of an aqueous solution of `NaOH(pH=12.0)` ?

Calculate the degree of hydrolysis of 0.1 M solution of sodium acetate at `298 K : K_(a) = 1.8 xx 10^(-5)`.

Neutral solutions have pH = 7 at 298 K. A sample of pure water is found to have pH `lt ` 7. Does it mean that it is acidic ? Explain.

The pH of 0.004 M hydrazine solution is 9.7. Calculate its ionization constant Kb and pKb.

Find the pH of the following solution:(a) 3.2 g of hydrogen chloride dissolved in 1.00 L of water.(b) 0.28 g of potassium hydroxide dissolved in 1.00 water.

The ionization constant of propionic acid is `1.32xx10^(-5)`. Calculate the degree of ionization of the acid in its `0.05`M solution and also its pH. What will be its degree of ionization in the solution of `0.01N HCI` ?

Solution of `0.1 N NH_(4)OH` and `0.1 N NH_(4)Cl` has `pH 9.25`, then find out `K_(b)` of `NH_(4)OH`.A. `9.25`B. `4.75`C. `3.75`D. `8.25`

The concentration of water molecules in pure water at 298 K isA. `10^(-7)M`B. `55.5M`C. `5.55M`D. `7.26M`

Determine the degree of ionization and pH of a 0.05 M of ammonia solution. The ionization constant of ammonia is 1.77 x 10-5 . Also calculate the ionization constant of conjugate acid of ammonia.

What is the hydrogen ion concentration of a solution with pH value 5.6 ?

Determine the degree of ionisation and pH of a 0.05 M ammonia solution. Also calculate ionisation constant of the conjugate acid ammonia. Given that `K_(b)` for `NH_(3)" is "1.77 xx 10^(-5)`.

For the reaction `CO(g) +H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)` at a given temperature the equilibrium amount of `CO_(2)` (g) can be increased byA. adding a suitable catalystB. adding an inert gasC. decreasing the volume of the containerD. increasing the amount of CO (g)

The `pH` of `0.05M` aqueous solution of diethy`1` amine is `12.0` . Caluclate `K_(b)`.

Write the conjugate acid for the following (a) Cl–, (b) NO3–, (c) HSO4–, (d) HCO3–, (e) SO42-, (f) CO32-, (g) NH2-, (h) N3- , (i) O2-, (j) H2O, (k) NH3.

What is the conjugate acid and conjugate base of HCO3 - ion ?

Write the conjugate base of HCO3

Calculate the pH of 0.001 NHCl.

The pH of a solution on addition of 50 ml. water slowly to a solution(a) 1(b) 5(c) 7(d) 10

At `450 K, K_(p)=2.0xx10^(10)//` bar for the given reaction at equilibrium. `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)` What is `K_(c )` at this temperature?

The reaction `2SO_(2)+2O_(2)hArr 2SO_(3)` will be favoured byA. high temperature and low pressureB. low temperature and high pressureC. high temperature and high pressureD. low temperature and low pressure.

What is chemical equilibrium ?

Write any two factors which affect ionization.

What are reversible and irreversible reactions?

What is the effect of adding an inert gas a) at constant volume b) at constant pressure on the following equation: PCl5 (g) PCl3(g) +Cl2(g)

which of the following reactions will get affected by increasing pressure? Mention whether the change will cause the reaction to go into forward or backward direction.(i) COCl2 (g) ----&gt; CO (g) + Cl2(g) (ii) CO2 (g) +C(s) -----&gt; 2CO (g) (iii) 2H2 (g) +CO (g) -------&gt; CH3OH (l)

Which of the following reaction will not affected on increasing the pressure ?A. `2H_(2(g))+CO_((g))hArrCH_(3)OH_((g))`B. `4NH_(3(g))+5O_(2(g))hArr4NO_((g))+6H_(2)O_((g))`C. `CH_(4(g))+2S_(2(g))hArrCS_(2(g))+2H_(2)S_((s))`D. `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`

Ionisation constant of `CH_(3)CO O H ` is `1.7xx10^(-5)` and concentration of `H^(+)` ions is `3.4 xx 10^(-4)`. Then initial concentration of `CH_(3) C O O H` molecules isA. `3.4xx10^(-4)`B. `3.4xx10^(-3)`C. `6.8xx10^(-3)`D. `1.7xx10^(-3)`

In the following reactions, ZnO is respectively acting as a/an `(i) ZnO + Na_(2)O rarr Na_(2) ZnO_(2)` `(ii) NzO + CO_(2) rarr ZnCO_(3)`A. Acid and acidB. Acid and baseC. Base and acidD. Base and base

Solubilirty product constants `(K_(sp))` of salts of types `MX, MX_(2)`, and `M_(3)X` at temperature T are `4.0xx10^(-8), 3.2xx10^(-14)`, and salts of temperature T is in the orderA. `MX_(2)gtM_(3)XgtMX`B. `M_(3)XgtMX_(2)gtMX`C. `MXgtM_(3)XgtMX_(2)`D. `MXgtMX_(2)gtM_(3)X`

A solution has `pH=5`, it is diluted `100` times, then it will becomeA. neutralB. basicC. unaffectesD. more acidic

`2mol` each of `A` and `B` are taken in a container to carry out the following reaction: `2A(g)+B(g)hArr2C(g)+2D(g)` When the system attains equilibrium, we haveA. `[A]lt[B]`B. `[A]gt[B]`C. `[A]=[B]`D. `[A]=[B]=[C]=[D]`

Which of the following anions is the weakest base ?A. `C_(2)H_(5)O^(-)`B. `CN^(-)`C. `CH_(3)COO^(-)`D. `NO_(3)^(-)`

How does common ion affect the solubility of electrolyte?

Equilivalent amounts of `H_(2)` and `I_(2)` are heated in a closed vessel till equilibrium is obtained. If `80%` of the hydrogen is converted to `HI`, the `K_(c)` at this temperature isA. 64B. 16C. `0.25`D. 14

which of the following reactions will get affected by increasing pressure? Mention whether the change will cause the reaction to go into forward or backward direction.(i) COCl2 (g) ----> CO (g) + Cl2(g) (ii) CO2 (g) +C(s) -----> 2CO (g) (iii) 2H2 (g) +CO (g) -------> CH3OH (l)