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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. |
Solubility (s) of `CaF_(2)` in terms of its solubility product is given asA. `s = (K_(sp))^(1//3)`B. `s = (K_(sp)//2)^(1//3)`C. `s = (K_(sp)//4)^(1//3)`D. `s = (K_(sp)//2)^(1//2)` |
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Answer» Correct Answer - C It is ternary electrolyte, `CaF_(2) harr Ca^(2+) + 2F^(-)` Let solubility of `CaF_(2)` be s mol `"litre"^(-1)` Then `{:(CaF_(2) hArr,Ca^(2+),+,2F^(-)),(,s,,2s):}` Let `K_(sp) = [Ca^(2+)] [F^(-)]^(2) = (s) xx (2s)^(2) = 4s^(3)` `s = sqrt((K_(sp))/(4))`. |
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| 252. |
At `25^(@)`C, `K_(sp)` for `PbBr_(2)` is equal to `8xx10^(-5)`. If the salt is `80%` dissociated, What is the solubility of `PbBr_(2)` in mol//litre?A. `[(10^(-4))/(1.6 xx 1.6)]^(1//3)`B. `[(10^(-5))/(1.6 xx 1.6)]^(1//3)`C. `[(10^(-4))/(0.8 xx 0.8)]^(1//3)`D. `[(10^(-3))/(1.6 xx 1.6)]^(1//2)` |
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Answer» Correct Answer - A `PbBr_(2(s)) hArr underset(0.8(s))(Pb_((aq))^(+2))+ underset(2 xx 0.8(s))(2Br_((aq))^(-))` |
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| 253. |
If the solubility of `Pbbr_(2)` is S g-mole per litre, its solubility product, considering it to be 80% ionized, isA. `2.048 S^(2)`B. `20.48 S^(3)`C. `2.048 S^(3)`D. `2.048 S^(4)` |
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Answer» Correct Answer - C `PbBr_(2) hArr Pb^(2+) + 2Br^(-)` `S (S xx 80)/(100) (2S xx 80)/(100)` [`because PbBr_(2)` ionises to 80%] `K_(sp) = [(S xx 80)/(100)][(2S xx 80)/(100)]^(2)` `= 0.8 S xx 2.56 S^(2) = 2.048 S^(3)` |
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| 254. |
The `K_(sp)` of `AgC1` at `25^(@)C` is `1.6 xx 10^(-9)`, find the solubility of salt in `gL^(-1)` in water. |
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Answer» Correct Answer - A::C::D `{:(AgC1rarr,Ag^(o+)+,C1^(Theta),,),(,S,S,,):}` `K_(sp) = S^(2)` `S = sqrt(K_(sp)) = (1.6 xx 10^(-9))^((1)/(2))` `= (16 xx 10^(-10))^((1)/(2)) = 4 xx 10^(-5)M` `("Mw of" AgC1 = 143.5)` Thus, `S = 4 xx 10^(-5) xx 143.5 = 5.74 xx 10^(-3) gL^(-1)` |
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| 255. |
`Pb(IO_(3))_(2)` is a sparingly soluble salt `(K_(sp) = 2.6 xx 10^(-13))`. To `35mL` of `0.15M Pb(NO_(3))_(2)` solution, `15mL` of `0.8M KIO_(3)` solution is added, and a precipiatte of `Pb(IO_(3))_(2)` is formed. Which is the limiting reactant of teh reaction that takes place in the solution?A. `Pb(IO_(3))_(2)`B. `Pb(NO_(3))_(2)`C. `KIO_(3)`D. Both (b) and (c). |
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Answer» Correct Answer - B `{:(2KIO_(3)+,Pb(NO_(3))_(2)rarr,Pb(IO_(3))_(2)+,2KNO_(3),),(0.8xx15,35xx0.15,,,),(=12,=5.25,,,):}` 1 mol of `Pb(NO_(3))_(2)` reacts with 2 mol of `KIO_(3)` `5.25` mmol of `Pb(NO_(3))_(2)` reacts `= 2 xx 5.25 = 10.5` mmol mmol of `KIO_(3)` left `= 12 - 10.5 = 1.5` Hence `Pb(NO_(3))_(2)` is the limiting reactant. |
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| 256. |
The solubility of `Mg (OH)_(2)` in pure water is `9.57 xx 10^(-3) gL^(-1)`. Calculate its solubility (in `gL^(-1))` in `0.02M Mg(NO_(3))_(2)` solution.A. `1.5 xx 10^(-4)`B. `8.69 xx 10^(-4)`C. `0.5 xx 10^(-3)`D. `0.5 xx 10^(-5)` |
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Answer» Correct Answer - B Solubility of `Mg(OH)_(2)` `= (9.57 xx 10^(-3))/(58) = 1.65 xx 10^(-4) "mol lits"^(-1)` `K_(sp) = 4s^(3)` `= 4xx (1.65 xx 10^(-4))^(3)` `= 17.96 xx 10^(-12)` But in presence of `Mg(NO_(3))_(2)` `K_(s) = (x+c)(2s)^(2)` `K_(s) = 17.96 xx 10^(-12) = (s+0.02)(2s)^(2)` `s = 14.98 xx 10^(-6) "mollts"^(-1)` `s = 14.98 xx 10^(-6) xx 58 gr"lt"^(-1)` `8.69 xx 10^(-4) gr "lits"^(-1)` |
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| 257. |
`K_(sp)` of `PbBr_(2)` is `8 xx 10^(-5)`. If the salt is `80%` dissociated in solution, calculat the solubility of salt in `gL^(-1)`. |
| Answer» Correct Answer - `12.48g litre^(-1);` | |
| 258. |
Carbonic acid , `H_2CO_3` is a diprotic acid for which `K_1=4.2xx10^(-7)` and `K_2=4.7xx10^(-11)` Which solution will produce a pH closest to 9 ?A. 0.1 M `H_2 CO_3`B. 0.1 M `Na_2 CO_3`C. 0.1 M `NaHCO_3`D. 0.1 M `NaHCO_3 and 0.1 M Na_2CO_3` |
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Answer» Correct Answer - c |
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| 259. |
`Pb(IO_(3))_(2)` is a sparingly soluble salt `(K_(sp) = 2.6 xx 10^(-13))`. To `35mL` of `0.15M Pb(NO_(3))_(2)` solution, `15mL` of `0.8M KIO_(3)` solution is added, and a precipiatte of `Pb(IO_(3))_(2)` is formed. What will be the molarity of `IO_(3)^(Theta)` ions in the solution after completion of the reaction?A. `0.152`B. `0.081`C. `0.41`D. `0.03` |
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Answer» Correct Answer - D `[IO_(3)^(Theta)] = (1.5 "mmol")/((15+35) mL) = 0.03M1` |
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| 260. |
The solubility of `Mg (OH)_(2)` in pure water is `9.57 xx 10^(-3) gL^(-1)`. Calculate its solubility (in `gL^(-1))` in `0.02M Mg(NO_(3))_(2)` solution. |
| Answer» Correct Answer - A::D | |
| 261. |
In aqueous solution the ionizatin constants for carbonic acid are : `K_1=4.2xx10^(-7) and K_2=4.8xx10^(-11)` Select the correct statement for a saturated 0.034 M solution of the carbonic acid.A. The concentration of `CO_3^(2-)` is 0.034 MB. The concentration of `CO_3^(2-)` is greater than that of `HCO_3^-`C. The concentration of `H^+` and `HCO_3^-` are approximately equalD. The concentration of `H^+` is double that of `CO_3^(2-)` |
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Answer» Correct Answer - c |
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| 262. |
A 0.001 M ammonia solution is `5%` ionized the concentration of `OH^(-)` ion isA. `0.005 M`B. `0.0001 M`C. `0.0005 M`D. `0.05 M` |
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Answer» Correct Answer - C `[OH^(-)] = Calpha = C xx (%)/(100)` `= 0.01 xx (5)/(100) = 0.0005` |
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| 263. |
`K_(sp)` for lead iodate `[Pb(IO_(3))_(2) is 3.2 xx 10^(-14)` at a given temperature. The solubility in `molL^(-1)` will beA. `2.0 xx 10^(-5)`B. `(3.2 xx 10^(-7))^(1//2)`C. `3.8 xx 10^(-7)`D. `4.0 xx 10^(-6)` |
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Answer» Correct Answer - A `K_(sp)=[Pb^(+)]^(2)[IO_(3)^(-)]^(2)=[s(2s)^(2)]=4s^(3)` `s=3 sqrt((K_(sp))/(4))=3sqrt((3.2 xx 10^(-14))/(4))` `3sqrt(8 xx 10^(-15))=2xx10^(-5)mol L^(-1)` |
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| 264. |
`K_(sp)` for lead iodate `[Pb(IO_(3))_(2) is 3.2 xx 10^(-14)` at a given temperature. The solubility in `molL^(-1)` will beA. `2.0 xx 10^(-5)`B. `(3.2 xx 10^(-7))^(1//2)`C. `(3.8 xx 10^(-7))`D. `4.0 xx 10^(-6)` |
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Answer» Correct Answer - A `{:([Pb(IO_(3))_(2)]hArr,Pb^(2+)+,2IO_(3)^(Theta),),(,2S,S,):}` `K_(sp) = 4S^(3)`. `S =root3((K_(sp))/(4))root3((3.2xx10^(-14))/(4))=2xx10^(-5)M`. |
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| 265. |
The `K_(sp)` of `BaSO_(4)` is `1.6 xx 10^(-9)`. Find the solubility of `BaSO_(4)` in `gL^(-1)` in a. Pure water b. `0.1M Ba(NO_(3))_(2)` |
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Answer» Correct Answer - A::B::C i. `{:(BaSO_(4) hArr,Ba^(2+)+,SO_(4)^(2-),,),(,S,2S,,):}` `S_(H)_(2)O) = sqrt(K_(sp)) = (1.6 xx 10^(-9))^(1//2) = 4 xx 10^(-5)M` `("Mw of" BaSO_(4) = 137 + 32 + 64 = 233 g)` `S_(H_(2)O)"in" gL^(-1) = 4 xx 10^(-5) xx 233` `= 9.32 xx 10^(-3)gL^(-1)` ii. In pressure of `0.1M Ba(NO_(3))_(2)`, due to common ion effect, the solubility is supressed. For unti-univalent or di-divalent salt, `S_("new") = (K_(sp))/((C)^(n))` (where `C` is the concentration of common ion added and `n` is the number of common ion in the salt `(BaSO_(4))`. `S_("new") = (1.6 xx 10^(-9))/((0.1)^(1)) = 1.6 xx 10^(-8)M` `S_("new") "is" gL^(-1) = 1.6 xx 10^(-8)xx233` `= 3.72 xx 10^(-6)gL^(-1)`. |
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| 266. |
Aqueous solutions of `Na_(2)C_(2)O_(4)` and `CaCI_(2)` are mixed and precipitate of `CaC_(2)O_(4)` formed is filered and dried. `250 mL` of the saturated solution of `CaC_(2)O_(4)` required `6.0 mL` of `0.001M KMnO_(4)` solution in acidic medium for complete titration. `K_(sp)` of `CaC_(2)O_(4)` isA. `2.25xx10^(-12)`B. `2.25xx10^(-10)`C. `3.6xx10^(-9)`D. `4.0xx10^(-9)` |
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Answer» Correct Answer - C Moles of `C_(2)O_(4)^(2-) rarr 1.5 xx 10^(-5) (250 mL)^(-1)` `[C_(2)O_(4)^(2-)] = [Ca^(2+)] = (1.5 xx 10^(-5)xx1000mL)/(250mL) = 6 xx 10^(-5)M` `K_(sp) = [Ca^(2+)] [C_(2)O_(4)^(2-)] = (6 xx 10^(-5))^(2) = 3.6 xx 10^(-9)M^(2)` |
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| 267. |
A weak mono acidic base is `5%` ionized in `0.01 M` solution. The Hydroxide ion concentration in the solution isA. `5 xx 10^(-2)`B. `5 xx 10^(-4)`C. `5xx 10^(-10)`D. `2 xx 10^(-11)` |
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Answer» Correct Answer - B `[OH^(-)] = (%)/(100) xx c = (5)/(100) xx 0.01 = 5 xx 10^(-4)` |
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| 268. |
`1 c.c` of `0.1 N HCl` is added to `1` litres of `0.1 N NaCl` solution. The pH of the resulting solution will beA. 7B. 1C. 3D. 4 |
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Answer» Correct Answer - D `NaCl` is a neutral salt `V_(1)N_(1) = V_(2)N_(2)` `[OH^(+)] = (0.1 xx 1)/(1000)` |
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| 269. |
What is the solubility in pure water of `Ba(IO_(3))_(2)` in moles per litre at `25^(@)`C? [`K_(sp)(25^(@)C=6.0xx10^(-10)`]A. `1.2xx10^(-5)`B. `1.7xx10^(-5)`C. `5.3xx10^(-4)`D. `8.4xx10^(-4)` |
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Answer» Correct Answer - c |
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| 270. |
A sample of hard water contains `0.005` mole of `CaCI_(2)` per litre. What is the minimum concentration of `Na_(2)SO_(4)` (in mole/L) which must be exceeded for removing `Ca^(+)` ions from this water sample ? The solubility product of `CaSO_(4)` at `25^(@)C` is `2.4 xx 10^(-5)`. (Given your answer after multiplying with 10000 and excluding and decimal places) |
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Answer» Correct Answer - `0048` |
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| 271. |
A base is dissolved in water yields a solution with a hydroxide ion concentration of 0.05 mole `"litre"^(-1)`. The solution isA. BasicB. AcidC. NeutralD. Both (a) and (b) |
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Answer» Correct Answer - A `[OH^(-)]` ion conc. `0.05 (mol)/(l) = 5 xx 10^(-2) (mol)/(l)` `pOH = -log [OH^(-)] = -log[ 5xx 10^(-2)]` `pOH = 1.30, pH + pOH = 14`. `pH = 14 - pOH = 14 - 1.30 = 12.7`. |
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| 272. |
`10 ml` of `0.1 N HCl` is added to `990 ml` solution of `NaCl` the `P^(H)` of resulting solutionA. ZeroB. `2`C. `3`D. `10` |
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Answer» Correct Answer - C `P^(H) = -log(10^(-1)) = 1` |
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| 273. |
The number of moles of hydroxide `(OH^(-))` ion in 0.3 litre of 0.005 M solution of `Ba(OH)_(2)` isA. 0.005B. 0.003C. 0.0015D. 0.0075 |
| Answer» Correct Answer - B | |
| 274. |
A solution which is `10^(-3)M` each in `Mn^(2+), Fe^(2+), Zn^(2+)` and `Hg^(2+)` is treated with `10^(-16) M` sulphide ion. If `K_(sp)` of MnS, FeS, ZnS and HgS are `10^(-15), 10^(-23), 10^(-20)` and `10^(-54)` respectively, which one will precipitate firstA. FeSB. MgSC. HgSD. ZnS |
| Answer» Correct Answer - C | |
| 275. |
`10mL` of a strong acid solution of `pH=2.000` are mixed with `990 mL` of another strong acid solution of `pH=4.000`. The `pH` of the resulting solution will be `:`A. `4.002`B. `4.000`C. `4.200`D. `3.7` |
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Answer» Correct Answer - 4 `[H^(+)]` after mixing `=(10^(-2)xx10+10^(-4)xx990)/(1000)=(0.1+0.0990)/(1000)=(0.1990)/(1000)=1.99xx10^(-4)...pH=4-0.3=3.7` |
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| 276. |
Equal volumes of `1M HCI` and `1M H_(2)SO_(4)` are neutralised by `1M NaOH` solution and `x` and `y kJ//` equivalent of heat are liberated, respectively. Which of the following relations is correct?A. not a buffer solution and with `pHlt7`B. not a buffer solution with `pHgt7`C. a buffer solution with `pHlt7`D. a buffer solution with `pHgt7` |
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Answer» Correct Answer - A pH is less than than 7 due to HCI not buffer. |
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| 277. |
Fixed volume of `0.1M` benzoic acid `(pK_(a) = 4.2)` solution is added into `0.2M` sodium benzote solution and formed a `300mL`, resulting acidic buffer solution. If `pH` of the resulting solution is `3.9`, then added volume of banzoic acid isA. `240mL`B. `150mL`C. `100mL`D. None |
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Answer» Correct Answer - A Let `VmL` of acid solution is taken. mmol of acid `= 0.1 xx V` mmol of salt `= 0.2M xx (300 - V) mL` `pH = pK_(a) + "log" (["Salt"])/(["Acid"])` `3.9 = 4.2 +"log" (["Salt"])/(["Acid"])` `-0.3 = "log"(["Salt"])/(["Acid"])` `"log" (["Acid"])/(["Salt"]) = 0.3` `(["Acid"])/(["Salt"]) = "Antilog" (0.3) = 2` `:. ["Acid"] =2 ["Salt"]` `(0.1 xx V)/(cancel 300) = 2 xx [(0.2xx(300-V))/(cancel300)]` Solve for `V:` `V = 240 mL` |
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| 278. |
`20 cm^(3)` of `xM` solution of `HCI` is exactly neutralised by `40cm^(3)` of `0.05 M NaOH` solutions, the `pH` of `HCI` solution isA. 1B. 2C. 1.5D. 2.5 |
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Answer» Correct Answer - A Moles of NaOH `=0.05xx40xx10^(-3)` Moles of HCl `=20xx 10^(-3)xx x` `0.05 xx 40 xx 10^(-3)=20 xx x xx 10^(-3)` or `x=(0.05xx40)/(20)=0.1 ` or `1xx10^(-1)` `:. [HCl]=[H_(3)O^(+)]=1xx10^(-1)` `:. pH =-log (1xx10^(-1))=1` |
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| 279. |
pH of 0.1 M solution of a weak acid (HA) is 4.50. It is neutralised with NaOH solution to decrease the acid contant to half pH of the resulting solutinA. `4.50`B. `8.00`C. `7.00`D. `10.00` |
| Answer» Correct Answer - B | |
| 280. |
`20 cm^(3)` of `xM` solution of `HCI` is exactly neutralised by `40cm^(3)` of `0.05 M NaOH` solutions, the `pH` of `HCI` solution isA. `1.0`B. `2`C. `1.5`D. `2.5` |
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Answer» Correct Answer - A `N_(1)V_(1)(HCI) = N_(2)V_(2)(NaOH)` `x xx 20 = 40 xx 0.05` `x = 0.1` `:.M of HCI = 0.1`. `[H_(3)O^(o+)] = 0.1 = 10^(-1)`. `pH = 1` |
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| 281. |
When `0.2M` solution of acetic acid is neutralised with `0.2M NaOH` in `500 mL` of water, the `pH` of the resulting solution will be: `[pK_(a)` of acetic acid `= 4.74]`A. `12.67`B. `7.87`C. `8.87`D. `7` |
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Answer» Correct Answer - C `CH_(3)COOH +NaOH rarr CH_(3)COONA +H_(2)O` `rArr` At equivalence point, a salt of weak acid, strong base is formed. `rArr pH = 7 +(1)/(2)(pK_(a) +logC)` Here `C=` concentration of salt `= (0.2 xx 500)/(500+500) =0.1M` [Check that `500mL` of `CH_(3)COOH` is also required] `rArr pH 7+(1)/(2) (4.75 + log0.1) = 8.87` |
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| 282. |
How much sodium acetate should be added to a 0.1 M solution of `CH_(3)COOH` to give a solution of pH = 5.5 (`pK_(a)` of `CH_(3)COOH = 4.5`)A. 0.1 MB. 0.2 MC. `1.0 M`D. `10.0 M` |
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Answer» Correct Answer - C `pH = pK_(a) + log.(["Salt"])/("Acid") , 5.5 = 4.5 + log.(["Salt"])/("Acid")` `log.(["Salt"])/("Acid") = 5.5 - 4.5 = 1` `(["Salt"])/(0.1) = "antilog 1" = 10, ["Salt"] = 1` . |
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| 283. |
If 20 cc. of HCl solution is exactly neutralised by 40 cc of 0.05 N NaOH, then pH of HCl isA. `3.0`B. `1.5`C. `2.0`D. `1.0` |
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Answer» Correct Answer - D Normality of HCl `=(40 xx 0.05 )/(20) =0.1` `:. pH =1.0` |
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| 284. |
What is the pH of a 0.025 M solution of KOH ?A. `1.60`B. `3.69`C. `10.31`D. `12.40` |
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Answer» Correct Answer - d |
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| 285. |
A buffer solution contains 0.1 mole of sodium acetate in `1000 cm^(3)` of 0.1 M acetic acid. To the above buffer solutions, 0.1 mole of sodium acetate is further added and dissolved. The pH of the resulting buffer is equal to............A. `pK_(a) - log_(2)`B. `pK_(a)`C. `pK_(a) + 2`D. `pK_(a) + log 2` |
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Answer» Correct Answer - D `pH = pK_(a) + log.(["Salt"])/(["Acid"])` `["Salt"] = (0.1 + 0.1)/(1000 ml) = (0.2)/(1) = 0.2` `["Acid"] = (0.1)/(1000 ml) = (0.1)/(1) = 0.1` `pH = pK_(a) + log 2` `:. pH = 14 - 2 rArr pH = 12`. |
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| 286. |
Two buffer solutions, `A` and `B`, each made acetic acid and sodium acetate differ in their `pH` by one unit, `A` has satl: acid `=x: y`, has salt: acid `=y :x.` If `x gt y`, then the value of `x : y` isA. `10,000`B. `3.17`C. `6.61`D. `2.10` |
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Answer» Correct Answer - B Buffer `(A):` `pH_(1) =pK_(a) + log ((x)/(y)) …(i)` Buffer `(B):` `pH_(2) = pK_(a) + log ((x)/(y)) …(ii)` Since `x gty`. `:. pH_(1) - pH_(2) = 1 = "log" (x)/(y) -"log"(y)/(x)` `:. 1=2"log"(x)/(y)` `"log"(x)/(y) = (1)/(2) = 0.5` `(x)/(y) = "Antilog" (0.5) = 3.17`. |
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| 287. |
The pH of a solution at `25^(@)C` containing 0.10 m sodium acetate and 0.03 m acetic acid is (`pK_(a)` for `CH_(3)COOH = 4.57`)A. 4.09B. 5.09C. 6.1D. 7.09 |
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Answer» Correct Answer - B `pH = pK_(a) + log.(["salt"])/(["acid"]) = 4.57 xx + log.(0.10)/(0.03) = 5.09`. |
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| 288. |
At `25^(@)C`, pH of a `10^(-8)M` aqueous KOH solution will beA. `6.0`B. 7.01C. 8.02D. 9.02 |
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Answer» Correct Answer - B `[Obar(H)]_("Total") = (10^(-8) + 10^(-7)) M :. pOH = -log [10^(-8) + 10^(-7)]` `= ~ 6.98` `:. pH = 14 - 6.98 = 7.02` |
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| 289. |
Which of the following has the highest pHA. 0.1 M NaOHB. 0.01 M NaOHC. 0.1 M HClD. `0.1 M CH_(3)COOH` |
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Answer» Correct Answer - A Base have higher pH than acids, stronger is the conc. of base still higher is the pH. |
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| 290. |
The pH of a simple acetate buffer is given by `pH = pK_(a) + log.(["Salt"])/(["Acid"])` `K_(a)` of acetic acid `= 1.8 xx 10^(-5)` If [Salt] = [Acid] = 0.1 M, the pH of the solution would be aboutA. 7B. 4.7C. 5.3D. 1.4 |
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Answer» Correct Answer - B `["Salt"] = 0.1 M, ["Acid"] = 0.1 M` `K_(a) = 1.8 xx 10^(-5) , pH = -logK_(a) + log.(["Salt"])/(["Acid"])` `= -log 1.8 xx 10^(-5) + log.(0.1)/(0.1) = -log 1.8 xx 10^(-5)` `pH = 4.7`. |
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| 291. |
A buffer solution contains 0.1 M of acetic acid and 0.1 M of sodium acetate. What will be its pH, it `pK_(a)` of acetic acid is 4.75A. `4.00`B. 4.75C. `5.00`D. 5.25 |
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Answer» Correct Answer - B `pH = pK_(a) + log.(["Salt"])/(["Acid"]), pH = 4.75 + log.(0.1)/(0.1)` `pH = 4.75 + log 1, pH = 4.75`. |
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| 292. |
In acid-base titration react rapidly to neutralise each other. Equivalence point is a point at which the acid and the base (or oxidising agent and reducing agent) have beem added in equivalent quantities. The end point in the point at which the titration stops. since the purpose of the indicator is to stop the titration close to the point at which the acid and base were added in equivalent quantities, it is important that the equivalent point and the end point be as close as must change colour at a `pH` close to that of a solution of the salt of the acid base. Singificantly, the `pH` changes most rapidly near the equivalent point. The exact shape of a titration curve depends on `K_(a)` and `K_(b)` of acid and base. Which of the following curves indicates the titration of a weak diprotic acid by `KOH` of equivalent strength?A. B. C. D. |
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Answer» Correct Answer - A In Fig (a), there are two equivalence point and `pH` changes most repidly at that point, so it is a titration of a weak diprotic acid by a strong base. At the end `pH` is very high and suggest a strong base. Figs. (b), (c) and (d) do not have high, `pH` at the end. So strong base is not possible. Moreover `pH` does not change repidly at equivalence point. |
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| 293. |
`1 xx 10^(-3)` mole of HCl is added to a buffer solution made up of 0.01 M acetic acid and 0.01 M sodium acetate. The final pH of the buffer will be (given `pK_(a)` of acetic acid is 4.75 at `25^(@)C`)A. `4.60`B. 4.66C. 4.75D. 4.8 |
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Answer» Correct Answer - B `{:(CH_(3)COO^(-),+,H^(+),rarr,CH_(3)COOH),(" "0.01,,0.001,," "0.01),(0.01-0.001,,,,0.01+0.001),(= 0.009,,,,= 0.011):}` `pH = pKa + log.(["salt"])/(["acid"]) = 4.75 + log. (0.009)/(0.011) = 4.66`. |
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| 294. |
In acid-base titration react rapidly to neutralise each other. Equivalence point is a point at which the acid and the base (or oxidising agent and reducing agent) have beem added in equivalent quantities. The end point in the point at which the titration stops. since the purpose of the indicator is to stop the titration close to the point at which the acid and base were added in equivalent quantities, it is important that the equivalent point and the end point be as close as must change colour at a `pH` close to that of a solution of the salt of the acid base. Singificantly, the `pH` changes most rapidly near the equivalent point. The exact shape of a titration curve depends on `K_(a)` and `K_(b)` of acid and base. The following curve represents titration curve of `HCI` against `KOH`. The `pH` at equivalent point is Examine the titration curve below and answer the question. A. `3`B. `6`C. `7`D. `8` |
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Answer» Correct Answer - C `pH` of titration of strong acid with strong base is `7`. |
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| 295. |
In acid-base titration react rapidly to neutralise each other. Equivalence point is a point at which the acid and the base (or oxidising agent and reducing agent) have beem added in equivalent quantities. The end point in the point at which the titration stops. since the purpose of the indicator is to stop the titration close to the point at which the acid and base were added in equivalent quantities, it is important that the equivalent point and the end point be as close as must change colour at a `pH` close to that of a solution of the salt of the acid base. Singificantly, the `pH` changes most rapidly near the equivalent point. The exact shape of a titration curve depends on `K_(a)` and `K_(b)` of acid and base. The curve represents the titration ofA. `CsOH by HBr`B. `HCI by NaOH`C. `HCI by KOH`D. `NH_(3) by HNO_(3)` |
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Answer» Correct Answer - A Since the curve starts at higher `pH (12.5)`, so it should be strong base and `pH` changes most rapidly near the equivalence point, so it should be strong acid. Hence the titration is strong base with strong acid (i.e., `CsOH` by `HBr)`. |
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| 296. |
Consider a buffer solution containing `0.1mol` each of acetic and sodium acetate in `1.0L` of solution, `0.01mol` of `NaOH` is gradully added to this buffer solution. Calculate the average buffer capcity of the solution and as well as initial and final buffer capcity. `[K_(a) = 2 xx 10^(-5)] pK_(a) = 4.7` |
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Answer» `pH` of the solutions is given by `pH = pK_(a_("initial")) +"log"_(10) (["salt"])/(["acid"]), pH_("initial") = 4.7` when `0.01M NaOH` is added. `["salt"] = 0.1 + 0.01 = 0.11M` and `["acid"] = 0.1 - 0.01 = 0.09M` `rArr` Final `pH` of solution: `pH = pK_(a) + "log" (0.11)/(0.09) = 4.787` `rArr [H^(o+)] = K_(a) (["acid"])/(["salt"])` `= 2 xx 10^(-5) xx (0.09)/(0.11) = 1.64 xx 10^(-5)M` `Delta(pH) = 4.787 - 4.7 = 0.087` Thus, `beta_(Avg) = (Deltan)/(Delta(pH)) = (0.01)/(0.087) = 0.1148` `C_("Buffer") = [HA] + [A^(Θ)] = 0.1 + 0.1 = 0.2M` `pH "initial" = 4.7, [H^(o+)] "initial" = 2 xx 10^(-5), K_(a) = 2 xx 10^(-5)` `beta_("initial") = 2.303 (C_("Buffer").K_(a)[H^(o+)])/([K_(a) +[H^(o+)]^(2))` `beta_("initial") = 2.303 (0.2 xx 2xx10^(-5) xx2xx10^(-5))/((2xx10^(-5)+2xx10^(-5))^(2)) = 0.115` `[H^(o+)]_("Final") = 1.64 xx 10^(-5)M`. `beta_("Final") = 2.303 (0.2 xx2xx10^(-5)xx1.64xx10^(-5))/((2xx10^(-5)+1.64xx10^(-5))^(2)) = 0.114` Note: Buffer capcity will decrease as more and more `NaOH` is added into the above solution. check yurself that initially the buffer capcity was maximum. so when `0.01M NaOH` was added, we will see a very slight decrease in the buffer capcity but as more and more `NaOH` is added, buffer capcity will decrease sharply and the difference between the average and maximum buffer capcity will get magnified. If `0.05M NaOH` is added in teh solution initially, we can calculate the final `pH` as: `pH = pK_(a) + log (0.15)/(0.05) = 5.177` Thus, `beta_(Avg.) = (Deltan)/(Delta(pH)) = (0.05)/(0.477) = 0.105` Calculate `beta_("initial")` and `beta_("Final")` yourself. |
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| 297. |
A buffer solution is formeed by mixing `100 mL` of `0.1 M CH_(3)COOH` with `200 mL` of `0.02 M CH_(3)COONa`. If this buffer solution is made to `1.0 L` by adding `700 mL` of water, pH will change by a factor of |
| Answer» No change in pH of buffer sol. By dilution | |
| 298. |
A buffer solution of `pH` value `4` is to be prepared, using `CH_(3)COOH` and `CH_(3)COONa`.How much amount of sodium acetate is to be added to `1.0L` of `M//10` acetic acid? `(K_(a) = 2.0 xx 10^(-5))` |
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Answer» For an acidic buffer containing aceetic acid, `CH_(3)COOH`, and sodium acetate, `CH_(3)COONa`, we have `[H^(o+)] = (K_(a) ["acid"])/[("salt")]` (Use this formula, rather than) `pH = pK_(a) + "log" (["salt"])/(["acid"])` `[CH_(3)COOH] = 0.1M,` `[H^(o+)] = 10^(-4)` and let `[CH_(3)COONa] = xM` `rArr [H^(o+)] = 10^(-4) = (2xx10^(-5)xx0.1)/(x)` `rArr x = 0.02` mol, i.e. `0.02` mol of `CH_(3)COONa` is required. |
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| 299. |
The pH of a buffer solution containing 25 ml of `1 M CH_(3)COONa` and 25 ml of `1M CH_(3)COOH` will be appreciably affected by 5 ml ofA. `1 M CH_(3)COOH`B. `5 M CH_(3)COOH`C. `5 M HCl`D. `1 M NH_(4)OH` |
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Answer» Correct Answer - B When ratio of concentration of acid to salt is increased pH decrease. |
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| 300. |
A buffer solution was made by adding `15.0 g` of `CH_(3) COOH` and `20.5gCH_(3)COONa`. The buffer is diluted to `1.0L`. a. Calculate the `pH` of solution. b. What will be the change in `pH` if `10.0mL` of `1.0 M HC1` is added to it. Given: `pK_(a) of CH_(3)COOH = 4.74, log ((13)/(12)) = 0.035` |
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Answer» Correct Answer - A::B::C::D i. Mw of `CH_(3)COOH = 60 g mol^(-1)` Mw of `CH_(3)COONa = 82g mol^(-1)` `["Salt"] = (20.5)/(82) = 0.25M (V = 1L)` `["Acid"] = (15.5)/(60) = 0.25M (V = 1L)` `pH = pK_(a) + log.(["Salt"])/(["Acid"])` `= 4.74 + log.([0.25])/([0.25]) = 4.74` ii. Rule `A A A:` In acidic buffer `(A)` on adding of `S_(A)(A)`, the concentration of `W_(A)(A)` increases and that of salt decrease. Moles of `HC1 = (10xx1M)/(1000) = 10^(-2)M = 0.01M` `[Acid] = 0.25 + 0.01 = 0.26M` `["Salt"] = 0.25 - 0.01 = 0.024` `pH_("new") = pK_(a) + "log" [("Salt")/("Acid")] = pK_(a) + log ((0.24)/(0.26))` `= pK_(a) + log ((12)/(13)) = pK_(a) - log ((13)/(12))` `= 4.74 - 0.035 = 4.705` Change in `pH:` `DeltapH = pH_("new") - pH_("inital")` `= 4.705 - 4.74 =- 0.035` |
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