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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
Which of the following reactions carried out in closed vessels are reversible ? (1) `2KCIO_(3)rarr 2KCI+O_(2)` (2) `N_(2)+O_(2)rarr2NO` (3) `PCI_(5)rarrPCI_(3)+CI_(2)` (4) `Fe^(3+)+SCN^(-)rarr[Fe(SCN)]^(2+)`A. (i), (ii), (iii), (iv)B. (ii), (iii), (iv)C. (i), (ii), (iii)D. (i), (iii), (iv) |
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Answer» Correct Answer - B Reaction (i) is irreversible because `O_(2)` cannot react with KCl to from `KClO_(3)`. |
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| 252. |
The strength of elctrolytes is expressed in terms of degree of dissociation `alpha` For strong electrolyte `alpha` is close to one and for weak electrolytes `alpha` is quite small. According to Ostwald Dilution Law `alpha=sqrt((K)/(C))` For an acid `[H^(+)] = sqrt(K_(a)C)` For a base `[OH^(-)] =sqrt(K_(b)C` The relative strengths of acids or bases can be compared in terms of the square roots of their `K_(a) " or " K_(b)` values. At infinite dilution , the percentage ionisation of both strong and weak electrolytes is :A. `1%`B. `20%`C. `50%`D. `100%` |
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Answer» Correct Answer - D Both strong and weak acids are almost completely ionised at infinite dilution. |
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| 253. |
The hydrogen ion concentration of a `10^(-8) M HCl` aqueous soultion at `298 K(K_(w)=10^(-14))` isA. `9.525xx10^(-8)M`B. `1.0xx10^(-8)M`C. `1.0xx10^(-6)M`D. `1.0525xx10^(-7)M` |
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Answer» Correct Answer - D HCI is a strong monobasic acid. Thus, `C_(H)^(+)` from HCl is equal to `C_(HCl)`, i.e., `10^(-8)` M. If we just focus on HCl, then pH of the solution will be 8: `C_(H^(+))=10^(-pH) mol L^(-1)` This is not acceptable as the solution is acidic. Its pH must be below 7. For this purpose, we must consider the contribution of `H^(+)` ions from water. In the absence of HCl, `C_(H^(+))` from water (at `25^(@)C`) is `10^(-7)` M. But due to the addition of HCl, the water equilibrium is distrubed and shifts backwards. Let the concentration of `H^(+)` ions be now x mol `L^(-1)` `H_(2)OhArr underset(x)(H^(+))+underset(x)(H^(-))` Total concentration of `H^(+)` ions will then be `(10^(-8)+x)` M. Thus, `K_(w)=C_(H^(+))C_(OH^(-))` `10^(-14)=(10^(-8)+x)(x)` or `x^(2)+10^(-8)x-10^(-14)=0` It is a quardatic equation `(ax^(2)+bx+c=0)` having the solution `x=(-b+-sqrt(b^(2)-4ac))/(2a)` `:. x= (-10^(-8)+-sqrt(10^(-16)+4xx10^(-14)))/(2)` `=9.51xx10^(-8)` This implies that `C_(OH^(-))=9.51xx10^(-8)` `:. C_(H^(+))=(10^(-14))/(C_(OH^(-)))=(10^(-14))/(9.51xx10^(-8))` `=1.05xx10^(-7)M` Also note that the only concentration which gives pH close to 7 but less than 7is `1.05xx10^(-7)`M. |
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| 254. |
Calculate the hydrogen ion and hydroxyl ion concentration of 0.01 M solution of `NaOH` at 298 K. |
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Answer» `NaOH` is a strong base and is , therefore, completely ionised in water. `{:(NaOH ,overset(aq)(to), Na^(+)(aq) ,+, OH^(-)(aq)),(,,0.01 M,,0.01M):}` `" Thus "" "[OH^(-) (aq)] =[NaOH]=0.01M` `"["H_(3)O^(+)"]" = (K_(w))/[[OH^(-)(aq)]]=((1.0xx10^(-14)M^(2)))/((0.01M)) = 10^(-12)M` |
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| 255. |
The equilibrium constant for the reaction : `Fe^(3) + (aq) +SCN^(-)(aq) hArr Fe SCN^(2+) (aq)` at 298 K is 138. What is the value of the equilibrium for the reaction? `2Fe^(3+) (aq) + 2SCN^(-) (aq) hArr 2Fe SCN^(2+) (aq)` |
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Answer» The expressions for the equilibrium constants of two reactions may be written as : `K_(1) = ([FeSCN^(2+) (aq)])/([Fe^(3+) (aq)][SCN^(-) (aq)]) = 138` `and " "K_(2) = ([FeSCN^(2+) (aq)]^(2))/([Fe^(3+) (aq)]^(2)[SCN^(-)(aq)]^(2))=[(FeSCN^(2+)(aq))/[[Fe^(3+) (aq)][SCN^(-)(aq)))]^(2)` `= (K_(1))^(2) = (138)^(2) = 19044` |
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| 256. |
The exact concentration of `H^(+)` ion in `10^(-3)` molar HCl aq solution at 298 K isA. `10^(-3)+10^(-7)`B. `10^(-3)+(K_(w))/([H^(+)])`C. `10^(-3)+(K_(w))/([OH^(-)])`D. `10^(-3)` |
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Answer» Correct Answer - B |
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| 257. |
Define ionic product of water. What is its value at 298 K? |
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Answer» It is defined as the product of molar concentration of hydrogen and hydroxyl ions. Kw=[H+][OH-] At 298 K , Kw = 10-14 M2 |
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| 258. |
The ionic product of water is 0∙11 x 10-14 at 273 K, 1 x 10-14 at 298 K and 7∙5 x 10-14 at 373K. Deduce from this data whether the ionization of water to hydrogen and hydroxide ion is exothermic or endothermic? |
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Answer» Kω = (H3O+)(OH−) According to the data, the value of Kω is increasing with temperature. Therefore, according to Le-Chatelier's principle, the ionisation of water is endothermic. |
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| 259. |
Write the correctly balanced net ionic equation for the reaction whose equilibrium constant at 298 K is:(i) ka (CaH5COOH) = 6.3 x 10-5(ii) ka (H2C2O4) = 5.4 x 10-2(iii) ka (HSO3-) = 2.8 x 10-7(iv) kb (CC1-) = 9.1 x 10-7(v) kb (CH3NH2) = 4.2 x 10-4(vi) ka (H2S) = 1.0 x 10-7(vii) ka (HCN) = 4.0 x 10-10(viii) kb (NH3) = 1.8 x 10-5(ix) ka (H2S) = 1.0 x 10-7 |
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Answer» (i) CaH5COOH ⇌ C6H5COO− + H+ (ii) H2C2O4 ⇌ C2O42− + 2H+ (iii) HSO3 ⇌ H+ + SO32- (iv) C1-H2O ⇌ OH- + HC1 (v) CH3NH2 + H2O ⇌ CH3NH3+ + OH- (vi) H2S ⇌ H+ + HS- (vii) HCN ⇌ H+ + CN- (viii) NH3 + H2O ⇌ NH4+ + OH- (ix) H2S + 2H2O ⇌ S2- + H3O+ |
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| 260. |
Match the following species with the corresponding conjugate acidSpeciesConjugate acid(i) NH3(a) CO32-(ii) HCO3-(b) NH4+(iii) H2O(c) H3O+(iv) HSO4-(d) H2SO4(e) H2CO3 |
| Answer» (i) → (b) (ii) → (e) (iii) → (c) (iv) → (d) | |
| 261. |
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant `K_(c)` is:A. `DeltaG=RTlnK_(c)`B. `-DeltaG=RTlnK_(c)`C. `DeltaG^(@)=RTlnK_(c)`D. `-DeltaG^(@)=RTlnK_(c)` |
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Answer» Correct Answer - D `-DeltaG^(@)=RTlnK_(c)` |
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| 262. |
Match standard free energy of the reaction with the corresponding equilibrium constant `{:((i) ,DeltaG^(Θ)gt0,,(a),K gt 1),((ii),DeltaG^(Θ) lt 0,,(b) ,K=1),((iii),DeltaG^(Θ) = 0 ,,(c),K=0),(,,,(d),K lt 1),(,,,,):}` |
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Answer» Correct Answer - (i)-(d), (ii)-(a), (iii)-(b) `DeltaG^(@)` = - RT ln K (i) If `DeltaG^(@) gt 0`, i.e., `Delta^(@)` is +ve, then ln K is -ve, i.e., `K lt 1` . Hence, (i)-(d). (ii) If `DeltaG^(@) lt 0`, i.e., `DeltaG^(@)` is -ve, then ln K is +ve, i.e., `K gt 1`. Hence, (ii) -(a). (iii) If `DeltaG^(@)= 0`, ln K = 1 , i.e., K = 1. Hence, (iii)-(b). |
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| 263. |
Why is equilibrium constant related to standard free energy change and not free energy change ? |
| Answer» Gibbs energy change is related to reaction quotient as `Delta _(r) G = Delta _(r) G^(@) + "RT In Q . When equilibrium is reached, " Delta_(r) G = 0 and Q = K. " Hence, we have " Delta _(r) G^(@)= -" RT In K ".` | |
| 264. |
The following reaction has attained equilibrium `CO(g) + 2 H_(2) (g) hArr CH_(3)OH (g) , Delta H^(@) = - 92*0 " KJ mol"^(-1)` What will happen if (i) volume of the reaction vessel is suddenly reduced to half ? (ii) the partial pressure of hydrogen is suddenly doubled ? (iii) an inert gas is added to the system ? |
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Answer» `K_(c) = ([ CH_(3)OH])/([CO][H_(2)]^(2)),K_(p)=(p_(CH_(3)OH))/(p_(CO)xx p_(H_(2))^(2))` (i) When volume of the vessel is reduced to Half, the concentration of each reactant or product becomes double. Thus, `Q_(c) = (2[CH_(3)OH])/(2[CO]xx{2[H_(2)]}^(2))=1/4 K_(c)` As ` Q_(c) lt K_(c)`, equilibrium will shift in the forward direction, producing more of `CH_(3)OH " to make " Q_(c) = K_(c)`. (ii) ` Q_(p) = (p_(CH_(3)OH))/p_(CO) xx (2 p_(H_(2)))^(2) = 1/4 K_(p).` Again , `Q_(p) lt K_(p).` equilibrium will shift in the forward direction to make `Q_(p) = K_(p)`. (iii) As volume remains constant, molar concentrations will not change. Hence there is no effect on the state of equilibrium. |
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| 265. |
The equilibrium `2Cu^(I) hArr Cu^(0) + Cu^(II)` in aqueous medium at `25^(@)C` shifts towards the left in the presence ofA. ` NO^(-)`B. `Cl^(-)`C. `SCN^(-)`D. `CN^(-)` |
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Answer» Correct Answer - B::C::D Cu (I) forms insouble compounds CuCl, CuCN and CuSCN. Hence , in the presence of `C1^(-) , CN^(-) and SCN^(-)` , the equilibrium shifts in the back direction . In fact, `Cu^(2+) " ions combine with " CH^(-) and SCN^(-)` to form compunds which are not stable and disociate to form `Cu^(+)` compounds/ complexes ` Cu^(2+) + 2 CN^(-) to Cu(CN)^(2)` ltbr ` Cu(CN)_(2) to 2 CuCN + (CN)_(2)` ` CuCN + 3 CN^(-) to [Cu (CN)_(4)]^(3-)` ` Cu^(2) + 4 SCN^(-) to [Cu(SCN)_(4) ]^(2-)` Similarly , `Cu^(2+) ` also combines with `Cl^(-)` ions to form `CuCl_(2)` which reacts with Cu to form CuCl, therby , shifting the equilibrium in the backward direction `CuCl_(2) + Cu to 2 CuCl` |
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| 266. |
At `0^(@)C`, ice and water are present in equilibrium. What will happen on increasing the pressure ? |
| Answer» On increasing the pressure, ice melts to form water (because water has lesser volume than ice). | |
| 267. |
The equilibrium ` SO_(2)Cl_(2) (g) hArr SO_(2) (g) + Cl_(2) (g) ` is attained at `25^(@)` in a closed container and inert gas helium is introduced . Which of the following statements are correct ?A. Concentration of `SO_(2) , Cl_(2)and SO_(2)Cl_(2)` changeB. More chlorine is formedC. Concentration of `SO_(2)` is reducedD. More `SO_(2)Cl_(2) ` is formed |
| Answer» No choice is correct because at constant volume, equilibrium is not disturbed . | |
| 268. |
The equilibrium `SO_(2) CI_(2) (g) hArr SO_(2)(g) +CI_(2)(g)` is attained at `25^(@)C` in a closed container and inert gas helium is introduced. Which of the following statement (s) is`//`are correct ? (1).concentrations of `SO_(2), CI_(2) " and " SO_(2) CI_(2)` change (2). More chlorine is formed (3).Concentration of `SO_(2)` is reduced (4).More `SO_(2) CI_(2)` is formedA. Concentration of `SO_(2), CI_(2) " and " SO_(2)CI_(2)` changeB. more of chlorine is formedC. Concentration of `SO_(2)` is reducedD. More of `SO_(2) CI_(2)` is formed. |
| Answer» None is correct because volume is constant. | |
| 269. |
The equilibrium `SO_(2) CI_(2) (g) hArr SO_(2)(g) +CI_(2)(g)` is attained at `25^(@)C` in a closed container and inert gas helium is introduced. Which of the following statement (s) is`//`are correct ? (1).concentrations of `SO_(2), CI_(2) " and " SO_(2) CI_(2)` change (2). More chlorine is formed (3).Concentration of `SO_(2)` is reduced (4).More `SO_(2) CI_(2)` is formedA. `1,2,3`B. `2,3,4`C. `3,4`D. None |
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Answer» Correct Answer - D When inert gas is added at constant volume there is not effect on equilibrium . Hence none of the options is correct answer. |
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| 270. |
Can equilibrium be attained in a reaction between acetic acid and ethyl alcohol carried in open container ? |
| Answer» Correct Answer - Yes | |
| 271. |
The equilibrium `H_(2) O (l) hArr H_(2)O (v) " is attained in a closed container at "40^(@)C.` The aqueous tension of water at `40^(@)C " is 23 mm . What is " K_(p)` for the said equilibrium ? |
| Answer» `K_(p) = p_(H_(2)O) (v)= 23 mm.` | |
| 272. |
Chemical equilibrium is attained in a reversible reaction carried in a close container and is of dynamic nature. The value of equilibrium constant may be expressed either as `K_(p)" and " K_(c)` and the two are related to each other as : `K_(p) =K_(c)(RT)^(Deltang)` Free energy change `(DeltaG)` at equilibrium point is zero. The value of equilibrium constant gives the extent to which a particular reation has proceeded to attain the equilibrium . Its value gets reversed if the reaction is reversed and becomes the square root of the initial value if the reaction is divided by 2. For the reaction `PCI_(3)(g)+ CI_(2) (s) hArr PCI_(5) (g)` The value of `k_(c) " at " 250^(@)C " is " " mol"^(-1) L^(-1)`. The value of `k_(p)` at the same temperature will be :A. `0.61 atm^(-1)`B. `0.56 atm^(-1)`C. `0.83 atm^(-1)`D. `0.46 atm^(-1)` |
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Answer» Correct Answer - A `K_(p) =K_(c)(RT)^(Deltang) =(26 "mol" L^(-1))` `x (0.0821 L atm K^(-1) mol^(-1) xx 523 K^(-1))` `= 0.61 K` |
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| 273. |
Chemical equilibrium is attained in a reversible reaction carried in a close container and is of dynamic nature. The value of equilibrium constant may be expressed either as `K_(p)" and " K_(c)` and the two are related to each other as : `K_(p) =K_(c)(RT)^(Deltang)` Free energy change `(DeltaG)` at equilibrium point is zero. The value of equilibrium constant gives the extent to which a particular reation has proceeded to attain the equilibrium . Its value gets reversed if the reaction is reversed and becomes the square root of the initial value if the reaction is divided by 2. A reaction attains equilibrium when the free energy change accompanying the reaction is :A. Positive and largeB. ZeroC. NegativeD. Negative and small. |
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Answer» Correct Answer - B `DeltaG=0` when the reaction is in a state of equilibrium |
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| 274. |
The reaction quotient (Q) for the reaction N2(g) + 3H2(g) ⇋ 2NH3(g) is given by Q = [NH3]2/[N2][H2]3. The reaction will proceed from right to left if(a) Q < Kc (b) Q > Kc(c) Q = 0 (d) Q = Kc |
| Answer» The correct answer is (b) Q > Kc | |
| 275. |
For the reaction, CO(g) + Cl2(g) ⇋ COCl2(g), the Kp/Kc is equal to(a) 1/RT (b) 1(c) (RT)-2 (d) RT |
| Answer» (a) The Kp/Kc is equal to 1/RT. | |
| 276. |
In a saturated solution of the sparingly soluble strong electrolyte `AgIO_(3)` (Molecular mass = 283), the equilibrium which sets in is `AgIO_(3) (s) hArr Ag^(+) (aq) +IO_(3)^(-) (aq)` If the solubility product constant `K_(sp)` of `AgIO_(3)` at a given temperature is `1.0xx10^(-8)`, what is the mass of `AgIO_(3)` contained in 100 ml of its saturated solution ?A. `1.0xx10^(-4)g`B. `28.3xx10^(-2)g`C. `2.83xx10^(-3)g`D. `1.0xx10^(-7)g` |
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Answer» Correct Answer - C `{:("AgIO"_(3),hArr,"Ag"^(+),+,"IO"_(3)^(-)),(,,S,,S):}` `K_(sp)=S^(2) or S=sqrt(K_(sp))=sqrt(10^(-8))=10^(-4) "mol" L^(-1)` `=10^(-4)xx283 g L^(-1) = 2.83 xx 10^(-2) g L ^(-1)` Thus, 1000 mL of the saturated solution will contain `AgIO_(3)=2.83xx10^(-2)g` `:. 100 ` m L of the saturated solution will contain `AgIO_(3)=2.83xx10^(-3)g`. |
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| 277. |
A sparingly soluble salt having general formula `A_(x)^(p+) B_(y)^(q-)` and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt. |
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Answer» Suppose molar solubility of `A_(x)^(p+) B_(y)^(q-)` is S mol `L^(-1)`. Then `{:(A_(x)^(p+)B_(y)^(q-) ,hArr,x A^(p+),+ ,yB^(q-),,),(,,x S,,y S,,):}` `K_(sp) = [ A^(p+)]^(x) [B^(q-)]^(y)=[x S]^(x) [ y S ] ^(y)= x^(x) y^(y) S^(x+y)` |
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| 278. |
Will AgCl be more soluble in aqueous solution or NaCl solutions and why ? |
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Answer» `AgCl hArr Ag^(+) + Cl^(-)` In NaCl solution, `[Cl^(-)]` will increase. As `[Ag^(+)][Cl^(-)]=K_(sp)` remains constant, `[Ag^(+)]` will decrease, i.e., the solubility will be less in NaCl solution than in water. |
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| 279. |
A saturated solution of sparingly soluble lead chloride on analysis was found to contain 11.84 g/ litre of the salt at room temperature. Calculate the solubility product constant at room temperature. (At. wt . : Pb = 207, Cl = 35.5 ) |
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Answer» Correct Answer - `3.09xx10^(-4)` Solubility of `PbCl_(2) = (11.84)/(207+2xx35.5) "mol" L^(-1) = 4.259 xx 10^(-2) "mol " L^(-1)` `PbCl_(2) rarr Pb^(2+)+2 Cl^(-), K_(sp)=[Pb^(2+)][Cl^(-)]^(2) = (4.259xx10^(-2))(2xx4.259xx10^(-2))^(2) = 3.09xx10^(-4)` |
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| 280. |
`PbCl_(2)` has a solubility product of `1.7xx10^(-8)`. Will a precipitate of `PbCi_(2)` form when 0.010 mole of lead nitrate and 0.010 mole of potassium chloride are mixed and water added upto 1 litre ? |
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Answer» Correct Answer - Yes As final volume of the solution = 1 L, therefore `[Pn^(2+)] = 0.01 M = 10^(-2) M, [ Cl^(_)] = 0.01 M = 10^(-2) M` Ionic product of `PbCl_(2) = [Pb^(2+)][Cl^(-)]^(2) (10^(-2))^(2) = 10^(_6)` which is greater than `K_(sp)`. Hence , ppt. of `PbCl_(2) ` will be formed. |
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| 281. |
NaCl solution is added to a saturated solution of `PbCl_(2).` What will happen to the concentration of `Pb^(+2)` ions? |
| Answer» `Pb^(+2)` ion concentration will decrease to keep `K_(sp)` constant. | |
| 282. |
NaCl solution is added to a saturated solution of `PbCl_(2)` . What will happen to the concentration of `Pb^(2+)` ions ? |
| Answer» `Pb^(+)` ion concentration will decrease to keep `K_(sp)` constant. | |
| 283. |
Calculate the solubility product of `PbCl_(2)` at a certain temperature if the solubility of `PbCl_(2)` is 4.4 g/L at the same temperature. `(Pb=207,Cl=35.5)` |
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Answer» Molar concentration of `PbCl_(2)=1.6xx10^(-2)M` `"[Concentration of"PbCl_(2)=4.4g//L=(4.4)/(278)=1.6xx10^(-2)mol//L"]"` `PbCl_(2)hArrPb^(+2)+2Cl` `{:(1,0,0),(1-s,s,2s):}` `K_(sp)=[Pb^(+2)][Cl^(-)]^(2)` `=sxx(2x)^(2)=4s^(3)` `=(1.6xx10^(-2))xx4=1.63xx10^(-5)` |
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| 284. |
Calcualte the solubility of `M_(2)X_(3)` in pure water, assuming that neither kind of ion reacts with `H_(2)O`. The solubility product of `M_(2)X_(3), K_(sp) = 1.1 xx 10^(-23)`. |
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Answer» `A_(2)X_(3)hArr2A^(+3)+3X^(-2)` `{:(1,0,0),(1-s,2s,3s):}` `K_(sp)=[A^(+3)]^(2)[X^(-2)]^(3)` `(2s)^(2)xx(3s)^(3)` `=4s^(2)xx27s^(3)` `=108s^(5)=1.1xx10^(-23)` `s=((K_(sp))/(108))^(1//5)` `=((1.1xx10^(-23))/(108))^(1//5)=10^(-5)mol L^(-1)` |
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| 285. |
Calcuate the degree of ionisation and pH of 0.05 M solution of a weak base having the ionization constant `(K_(b))` is `1.77xx10^(-5).` Also calculate the ionisation constnat of the conjugate acid of this base. |
| Answer» `alpha =1.88xx10^(-2),pH =10.96, k_(a)=5.64xx10^9-10` | |
| 286. |
One litre of `0.05` M HCl was completely neutralized by NaOH. Calcuate the pH of resulting solution |
| Answer» The resulting solution will be netural because strong acid is competely neturalized by storn base. Hence pH will be 7 | |
| 287. |
According to the law of chemical equilibrium,A. the rate of forward reaction becomes equal to the rate of backward reaction when the chemical system attains equilibriumB. a system can achieve the equilibrium state through forward as well as backward reactionC. both (1) and (2)D. the equilibrium constant `K_(eq)` is defined as the product of the equilibrium active masses of the products, each raised to the power that corresponds to its coefficient in the balanced equation, divided by the product of the equilibrium active masses of reactants, each raised to the power that corresponds to its coefficient in the balanced equation |
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Answer» Correct Answer - D For `ubrace(aA+bB)_("Reactants")hArr ubrace(cC+dD)_("Products")` The law of chemical equilibrium says `K_(eq)=([C]_(eq)^c[D]_(eq)^d)/([A]_(eq)^a[B]_(eq)^b)` |
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| 288. |
`K_f` and `K_b` are the velocity constants of forward and backward reactions. The equilibrium constant `K_(eq)` of the reversible reaction will beA. `K_b//K_f`B. `K_fxxK_b`C. `K_f//K_b`D. `K_f-K_b` |
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Answer» Correct Answer - C Suppose a reversible reaction occurs by a one-step mechanism: `2A+B=A_2B` The rate of the forward reaction is `Rate_f=K_f[A]^2[B]` The rate of the backward reaction is `Rate_b=K_b[A_2B]` At equilibrium, `Rate_f=Rate_b` `K_f[A]^2[B]=K_b[A_2B]` Dividing both sides of this equation by `K_b` and by `[A]^2[B]` gives `K_(eq)=K_f/K_b=([A_2B])/([A]^2[B])` |
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| 289. |
0.1 M `CH_(3)CO OH` (pH = 3) is titrated with 0.05 M NaOH solution. Calculate the pH when (i) 1/4th of the acid has been neutralized. (ii) 3/4th of the acid has been neutralized. |
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Answer» Calculation of dissociation constant of the acid `CH_(3)CO OH hArr CH_(3)CO O^(-) + H^(+)` As `pH = 3, :. [H^(+)] = 10^(-3)M, [CH_(3)CO O^(-)]=[H^(+)]=10^(-3)M` `K_(a) = ([CH_(3)CO O^(-)][H^(+)])/([CH_(3)CO OH])=(10^(-3)xx10^(-3))/(0.1) = 10^(-5)` (i) When 1/4th of the acid has been neutralized `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa ,+,H_(2)O),("Initial conc.",0.1 M,,,,,,),("After 1/4th neutralization",0.1xx(3)/(4),,,,0.1xx(1)/(4),,),(,=0.075 M ,,,,=0.025 M ,,):}` `:. pH = pK_(a) + log. (["Salt"])/(["Acid"])=-log 10^(-5) + log. (0.025)/(0.075) = 5 - 0.4771 = 4.5229` (ii) When 3/4th of the acid has been neutralized `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa , +,H_(2)O),("Initial conc.",0.1 M,,,,,,),("After 3/4th",0.1xx1/4M,,,,0.1xx3/4M,,),("neutralization",=0.025 M,,,,=0.075 M,,):}` `:. pH = - log 10^(-5) + log. (0.075)/(0.025) = 5 + 0.4771=5.4771` |
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| 290. |
Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH ?A. `SrCl_(2)`B. `BaCl_(2)`C. `MgCl_(2)`D. `CaCl_(2)` |
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Answer» Correct Answer - B All alkaline earth metal chlorides `(MCl_(2))` on hydrolysis will produce acidic solution because the base formed , M `(OH)_(2)`, is weak and acid formed (HCl) is strong. However, as we go down the group, basic character of hydroxides increases. Hence, acidic character decreases, i.e., pH increases (though it is `lt 7` in all cases ). Therefore, `BaCl_(2)` solution will have the highest pH . |
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| 291. |
Equimolar solution of the following were prepared in water separately. Which one of the solutions will record the highest `pH`?A. `SrCI_(2)`B. `BaCI_(2)`C. `MgCI_(2)`D. `CaCI_(2)` |
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Answer» Correct Answer - B Alkaline earth metal chlorides are salts of strong acid (HCI) and weak bases `[M(OH)_(2)]`. Thus, on hydrolysis, they basically yield acidic solution. As we go down the group, the basic character of hydroxides increases. This tends to reduce the acidic nature of solution, i.e., the pH of the resulting solution increases on moving down the group of alkaline earth metals. Thus, the solution of `BaCl_(2)` has the highest pH. |
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| 292. |
When a solution of benzoic acid was titrated with `NaOH` the `pH` of the solution when half the acid neutralized was `4.2`. Dissociation constant of the acid isA. `6.31xx10^(-5)`B. `3.2xx10^(-5)`C. `8.7xx10^(-8)`D. `6.42xx10^(-4)` |
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Answer» Correct Answer - A |
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| 293. |
The ionisation constant of benzoic acid `(PhCOOH)` is `6.46 xx 10^(-5)` and `K_(sp)` for silver benzoate is `2.5 xx 10^(-3)`. How many times is silver benzoate more soluble in a buffer of `pH 3.19` compared to its solubility is pure water? |
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Answer» (I) .Calculation of solubility in water `C_(6)H_(5)COOAg to C_(6)H_(5)COO^(-)+ Ag^(+)` Suppose solubility in water =x mol `L^(-1)` `[C_(6)H_(5)COO^(-)] =[Ag^(+)] =x mol L^(-1)` `x^(2) =K_(sp) " or " x(K_(sp))^(1//2) =(2.5 xx 10^(-13))^(1//2) = 5 xx 10^(-7) mol L^(-1)` (II). Calculation of solubiltiy in buffer of pH =3.19 `pH 3.19 " means " - log [H^(+)] =3.19` ` " or " " " log [H^(+)] =- 3.19 =bar(4).81 " or " [H^(+)] =6.457 xx 10^(-4)` `C_(6)H_(5)COO^(-)` ions now combine with the `H^(+)` ions to form benzoic acid but `[H^(+)]` remains almost constant because we have buffer solution. Now `C_(6)H_(5)COOH hArr C_(6)H_(5)COO^(-) +H^(+)` `:. =[[C_(6)H_(5)COO^(-)][H^(+)]]/[[C_(6)H_(5)COOH]] " or "[[C_(6)H_(5)COO^(-)]]/[[C_(6)H_(5)COOH]] =[[H^(+)]]/(K_(a)) =(6.457xx10^(-4))/(6.46xx10^(-5))=10` Suppose solubility in the buffer solution is y mol `L^(-1)`. Then as most of the benzoate ions are converted into benzoic acid molecules (which remain almost ionized), we have `y= [Ag^(+)] =[C_(6)H_(5)COO^(-)]+[C_(6)H_(5)COOH]` `=[C_(6)H_(5)COO^(-)]+ 10 [C_(6)H_(5)COO^(-)]=[C_(6)H_(5)COO^(-)]` `:. " "[C_(6)H_(5)COO^(-)]=(y)/(11)` `K_(sp) =[C_(6)H_(5)COO^(-)][Ag^(+)]` `i.e., " " 2.5 xx 10^(-13) =(y)/(1)) xx y " or "y^(2) =11 xx 2.5 xx 10^(-13)` `:. " " y^(2) =2.75 xx 10^(-12) " or " y = (2.75 xx 10)^(1//2)=1.66 xx 10^(-6) mol L^(-1)` `:. " "(y)/(x) =((1.66 xx 10^(-6) "mol " L^(-1)))/((5 xx 10^(-7) "mol " L^(-1))) = 3.32` |
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| 294. |
Prove that the pressure necessary to obtain 50% dissociation of `PCl_(5)` at 500 K is numerically three times the value of `K_(p)`. |
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Answer» `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Iitial molar conc".,1,,0,,0),("Eqm. molar conc".,(1-0.5),,0.5,,0.5):}` Total no. of moles at eqm. Point =1-0.5 +0.5 = 1.5 mol. `K_(p)= (pPCl_(3)xxpCl_(3))/(pPCl_(5))=(((0.5p)/(1.5))xx((0.5P)/(1.5)))/(((0.5P)/(1.5)) )= (0.5)/(1.5) P` `K_(p) =1/3 P or P = 3 K_(p)` |
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| 295. |
Prove that the pressure necessary to obtain 50% dissociation of `PCl_(5)` at 500 K is numerically equal to three times the value of the equilibrium constant, `K_(p)`. |
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Answer» ` {: (,PCl_(5),hArr,PCl_(3),+,Cl_(2),) , (" Intial moles",1,,0,,0,) , ("Moles at eqm".,1-0*5=0*5,,0*5,,0*5,"Total " 1*5 "moles") :}` If P is the total required pressure, then ` p_(PCl_(5)) = (0*5)/(1*5) xx P=P/3, p_(PCl_(3)) = (0*5)/(1*5) xx P = P/3, p_(Cl_(2)) = (0*5)/(1*5) xx P= P/3` ` K_(p) = (p_(pCl_(3)) xx p_(cl_(2)))/(p_(PCl_(5)))= ((P//3)(P//3))/(P//3) = P/3 or P = 3 K_(P).` |
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| 296. |
`K_(p) = 0*04` atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of `C_(2)H_(6)` when it is placed in a flask at 4*0 atm pressure and allowed to came in equilibrium ? ` C_(2) H_(6) (g) hArr C_(2) H_(4) (g) + H_(2) (g)` |
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Answer» ` {:(,C_(2)H_(6)(g),hArr,C_(2)H_(4)(g),+,H_(2)(g)),("Intial pressure",4*0"atm",,0,,0),("At eqm".,4-p,,p,,p):}` `K_(p) = (p_(C_(2)H_(4))xx p_(H_(2)))/(p_(C_(2)H_(6))):. 0*04 = p^(2)/(4-p) or p^(2) = 0*16 - 0*04 p ` or ` p^(2) + 0*04 p - 0*16 = 0 :. p = (-0*04 pm sqrt(0*0016 - 4(-0*16)))/2 = (0*04 pm 0*89)/2` aking positive value, ` p= (0*80)/2 = 0*40 :. [C_(2) H_(6) ]_(eq) = 4 - 0*40 "atm" = 3* 60 "atm " ` |
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| 297. |
Classify the following as acid or base according to Bronsted - Lowry concept. (i) `CH_(3)COO^(-)` (ii) `H_(3)O^(+)` (iii) `SO_(4)^(2-)` (iv) `HCl`A. `{:((i),(ii),(iii),(iv),),("Bronsted acid","Bronsted base","Bronsted base","Bronsted acid",):}`B. `{:((i),(ii),(iii),(iv),),("Bronsted acid","Bronsted acid","Bronsted acid","Bronsted base",):}`C. `{:((i),(ii),(iii),(iv),),("Bronsted base","Bronsted acid","Bronsted base","Bronsted acid",):}`D. `{:((i),(ii),(iii),(iv),),("Bronsted acid","Bronsted acid","Bronsted base","Bronsted base",):}` |
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Answer» Correct Answer - C Any species which can accept a proton is Bronsted base while which can give a proton is Bronsted acid. |
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| 298. |
Acidity of `BF_(3)` can be explained on ths basis of which of the follwoing concepts?A. Arrhenius conceptB. Bronsted - Lowry conceptC. Lewis conceptD. Bronsted - Lowry as well as Lewis concept |
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Answer» Correct Answer - C `BF_(3)` is an electron deficient compound and hence is Lewis acid. |
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| 299. |
Acidity of `BF_(3)` can be explained on the basis of which of the following concepts?A. Arrhenius conceptB. Bronsted Lowry conceptC. Lewis conceptD. Bronsted Lowry as well as Lewis concept. |
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Answer» Correct Answer - C `BF_(3)` is electron deficient compound and hence is a Lawis acid. |
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| 300. |
Acidity of `BF_(3)` can be explained on ths basis of which of the follwoing concepts?A. Arrhenius conceptB. Bronsted Lowry conceptC. Lewis conceptD. Bronsted Lowry as well as Lewis concept. |
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Answer» Correct Answer - C `BF_(3)` is a Lewis acid and can aceept a pair of electrons form a Lewis base. |
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