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InterviewSolution
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.
| 1401. |
The solubility product of barium sulphate is `1.5 xx 10^(-9)` at `18^(@)C` . Its solubility in water at `18^(@)C` isA. `1.5 xx 10^(-9)` mol `L^(-1)`B. `1.5 xx 10^(-5) mol L^(-1)`C. `3.9 xx 10^(-9) mol L^(-1)`D. `3.9 xx 10^(-5) mol L^(-1)` |
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Answer» Correct Answer - D `BaSO_(4) rarr Ba^(+2)+SO_(4)^(2-)` `:. ` Solubility `=(K_(sp))^(1//2)` `=(1.5 xx 10^(-9))^(1//2)=3.87 xx 10^(-5)mol L^(-1)` |
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| 1402. |
Out of the following salts identify the one which has maximum solubility (in M) : (P) Barium sulphate having `K_(sp)=1.1xx10^(-10)` (Q) Aluminium phosphate having `K_(sp) =6.3xx10^(-19)` (R ) Cadmium iodate having `K_(sp)=3.2xx10^(-8)` (S) Cuprous chloride having `K_(sp) =1.2xx10^(-6)`A. Aluminium phosphateB. Cuprous chlorideC. Barium sulphateD. Cadmium iodate |
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Answer» Correct Answer - d |
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| 1403. |
Which hydroxide will have lowest value of solubility product at normal temperature `(25^(@)C)`A. `Mg(OH)_(2)`B. `Ca(OH)_(2)`C. `Ba(OH)_(2)`D. `Be(OH)_(2)` |
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Answer» Correct Answer - D `Be(OH)_(2)` has lowest solubility and hence lowest solubility product. |
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| 1404. |
Which of the following species acts as Bronsted base but not as acidA. `HSO_(4)^(-)`B. `HCO_(3^(-))`C. `H_(2)PO_(2^(-))`D. `H_(2)PO_(3^(-))` |
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Answer» Correct Answer - C `H_(2)PO_(2)^(-)` is bronsted base. |
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| 1405. |
Which of the following sulphate has the lowest solubility product ?A. FeSB. MnSC. PbSD. ZnS |
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Answer» Correct Answer - C PbS has lowest solubility product of all the sulphides given in the question. `Pb^(2+)` cation belong to group -2 of salt analysis sulhides of group -2 have lower solubility products. |
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| 1406. |
`12g` of `CH_(3)COOH` and `4g` of `NaOH` are mixed and diluted to `1` litre solution. If `P^(Ka)` of `CH_(3)COOH` is `4.8` the `P^(H)` of the solution isA. `4.8`B. `5`C. `5.6990`D. `5.3010` |
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Answer» Correct Answer - A `CH_(3)COOH + NaOH rarr CH_(3)COONa + H_(2)O` `[CH_(3)COOH] = (0.2 - 0.1) - 0.1 = 0.1` `[CH_(3)COONa] = 0.1` `P^(H) = 4.8 + "log" (0.1)/(0.1)` `P^(H) = 4.8 + "log" (0.1)/(0.1)` `P^(H) = 4.8` |
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| 1407. |
1 litre of `1M CH_(3)COOH` (very weak acid) taken in a container initially. Now this solution is diluted upto volumen V (litre) so that pH of the resulting solution becomes twice the original value. `(K_(a) (CH_(3)COOH) = 10^(-6))`. Now equal volume of `0.5 xx 10^(-6)M NaOH` solution is added to this resulting solution, so that a buffer solution is obtained. Find `[H^(+)]` in original solutionA. `1 M`B. `10^(-3)M`C. `10^(-7)M`D. `10^(-11)M` |
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Answer» Correct Answer - b |
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| 1408. |
The `[H^(+)]` of 0.10 `MH_(2)S` solution is (given `K_(1)=1.0 xx 10^(-7) , K_(2)=1.0 xx 10^(-14))`A. `1.0 xx 10^(-7)M`B. `1.0 xx 10^(-4)M`C. `1.0 x 10^(-8)M`D. `1.0 xx 10^(-22)M` |
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Answer» Correct Answer - B As `K_(2) lt lt lt lt K_(1)` most of the `H^(+)` ions results from primary ionisation `H_(2)S hArr H^(+)+HS^(-)` Let `x=[H^(+)]=[HS^(-)]` `[H_(2)S]=1.0 -x=1.0 mol L^(-1)` ( Since x is very small ) `K_(1)=([H^(+)][HS^(-)])/([H_(2)S])` or `1.0 xx 10^(-7)=(x^(2))/(0.10)=1.0 xx 10^(-4)M` `[H^(+)]=x mol L^(-1)` `x=1.0 xx 10^(-4)` |
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| 1409. |
Find the `pH` of (a)`10^(-3)M HNO_(3)` solution (b)`10^(-4)MH_(2)SO_(4)` solution (Take`log2=0.3`) |
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Answer» `(a)p-log [H^(+)]_(HNO_(3))=-log(10^(-3))=3` `(b) pH=-log[H^(+)]_(H_(2)SO_(4))=-log(2xx10^(-4))=4-log2=3.7` In both solution,`[H^(+)]_("from strong acid") gt 10^(-6)M` so `H^(+)` from water has not been considered. |
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| 1410. |
For, `H_(3)PO_(4)+H_(2)OhArrH_(3)O^(+)+H_(2)PO_(4)^(-), K_(a_(1))` `H_(2)PO_(4)+H_(2)OhArrH_(3)O^(+)+HPO_(4)^(2-), K_(a_(2))` `HPO_(4)^(2-)+H_(2)OhArrH_(3)O^(+)+PO_(4)^(3-), K_(a_(3))` The correct order of `K_(a)` values is:A. `K_(a_(1))gtK_(a_(2))ltK_(a_(3))`B. `K_(a_(1))ltK_(a_(2))ltK_(a_(3))`C. `K_(a_(1))gtK_(a_(2))gtK_(a_(3))`D. `K_(a_(1))ltK_(a_(2))gtK_(a_(3))` |
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Answer» Correct Answer - C The dissociation constant for different ateps dissociation of weak polyrotic acid decreases continuously, i.e., `K_(a_(1))gtK_(a_(2))gtK_(a_(3))` |
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| 1411. |
Successive dissociation constants of `H_(2)CO_(3)` are `K_(a_(1)) = 10^(-3)` and `K_(a_(2)) = 10^(-6)`. An aqueous solution contains `0.1 MH_(2)CO_(3)` and `0.1 M HCI` at `25^(@)C` If solution contains only `0.1 M H_(2)CO_(3)` then the concentration of `CO_(3)^(2-)` at equilibrium in the solution will be :A. `10^(-6) M`B. `10^(-12)M`C. `10^(-9)M`D. `10^(-12)M` |
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Answer» Correct Answer - a |
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| 1412. |
Successive dissociation constants of `H_(2)CO_(3)` are `K_(a_(1)) = 10^(-3)` and `K_(a_(2)) = 10^(-6)`. An aqueous solution contains `0.1 MH_(2)CO_(3)` and `0.1 M HCI` at `25^(@)C` The concentration of `CO_(3)^(2-)` at equilibrium in solution is: (approximately)A. `10^(-6)M`B. `10^(-12)M`C. `10^(-10)M`D. `10^(-8)M` |
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Answer» Correct Answer - d |
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| 1413. |
Ratio of `[HA^(+)]` in `1L` of `0.1M H_(3)A` solution `[K_(a_(1))=10^(-5),K_(a_(2))=10^(-8)K_(a_(3))=10^(-11)]`& upon addition of `0.1`mole `HCl` to it will be :A. `10`B. `100`C. `1000`D. `10,000` |
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Answer» Initial :`[HA^(2-)]=10(-8)M=K_(a_(2))` final:`K_(a_(1))xxK_(a_(2))=([H^(+)][HA^(2-)])/([H_(3)A]):.10^(-8)xx10^(-5)=([H^(+)]^(2)[HA^(2-)])/([H_(3)A])rArr[HA^(2-)]=10^(-12)M` So ratio`=(10^(-8))/(10^(-12))=10^(4)` |
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| 1414. |
If `K_(a_(1)),K_(a_(2))` are the first, second and third ionization constants of `H_(2)PO_(4)` respectively and `K_(a_(1)) gtgt K_(a_(2)) gtgt K_(a_(3))`. Which is/are correct:A. `[H^(+)] = sqrt(k_(a_(1))[H_(3)PO_(4)])`B. `[H^(+)] = [HPO_(4)^(2-)]`C. `K_(a_(2)) = [HPO_(4)^(2-)]`D. `K_(a_(1)) = [HPO_(4)^(2-)]` |
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Answer» Correct Answer - A::C In case of polyprotic acids, divalent anion concentration is equal to `K_(a_(2))` |
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| 1415. |
For preparing a buffer solution of `pH 6` by mixing sodium accetate and acetic, the ratio of the concentration of salt and acid should be `(K_(a)=10^(-5))`A. `1:10`B. `10:1`C. `100 :1`D. `1:100` |
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Answer» Correct Answer - B `pH=pK_(a)+log. (["Salt"])/(["base"])` `pK_(a)=-log K_(a)=-log10^(-5)= +5` `log.(["Salt"])/(["base"])=pH -pK_(a)=6-5=1` `(["Salt"])/(["base"])=` antilog `(1)=10` Thus [salt]`:`[base]`10:1` |
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| 1416. |
At `20^(@)C`, the `Ag^(+)` ion concentration in a saturated solution `Ag_(2)CrO_(4)` is `1.5 x10^(-4)` mol `//` litre. At `20^(@)C`, the solubility product of `Ag_(2)CrO_(4)` would beA. `3.3750 xx 10^(-12)`B. `1.6875 xx 10^(-10)`C. `1.6875 xx 10^(-12)`D. `1.6875 xx 10^(-11)` |
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Answer» Correct Answer - C `Ag_(2)CrO_(4) hArr 2Ag^(+)+CrO_(4)^(-2)` Given `[2Ag^(+)]=1.5 xx 10^(-4) mol L^(-1)` Thus, `[Ag^(+)]=0.75xx 10^(-4) mol L^(-1)` Thus `[CrO_(4)^(-2)]=0.75 xx 10^(-4) mol L^(-1)` `K_(sp)=[Ag^(+)]^(2) [CrO_(4)^(-2)]^(2)` `=(1.5 xx 10^(-4))^(2)(0.75 xx 10^(-4))` `=1.6875xx 10^(-12)mol^(3)L^(-3)` |
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| 1417. |
In a solution containing `0.1M HCl & 0.1M H_(3)PO_(4)` (having dissociation constants `K_(a_(1)),K_(a_(2)),K_(a_(3)))` assuming `alphaltlt1` for `H_(3)PO_(4)`,A. `pH=1`B. `alpha_(H_(3)PO_(4))=10K_(a_(1))`C. `[HPO_(4)^(2-)]=10K_(a_(1))K_(a_(2))`D. All of these |
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Answer» `[H^(+)]=0.1M`& `pH=1` `K_(a_(1))=(0.1[H_(2)PO_(4)^(-)])/(0.1)rArr [H_(2)PO_(4)^(-)]=K_(a_(1))rArrCalpha=K_(a_(1))rArr alpha=(K_(a_(1)))/(0.1)=10K_(a_(1))` `K_(a_(2))=(0.1[HPO_(4)^(-)])/(K_(a_(1)))rArr[HPO_(4)^(-)]=K_(a_(1))xxK_(a_(2))xx10` |
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| 1418. |
Which one is hard base ?A. `Ag^(+)`B. `Cr^(3+)`C. `I_(2)`D. `F^(-)` |
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Answer» Correct Answer - D Hard base is that anion which is small and diffuclt to polarise. |
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| 1419. |
What volume of `0.1M` sodium formate solution should be added to 50 mL of `0.05` M formic acid to produce a buffer solution of `pH= 4.0`? `(pK_(a)` of fomic acid `= 3.80)` |
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Answer» Let V mL of `0.1M HCOONa` be mixed to 50 mL of `0.05 M HCOOH`. `[.: [ ] = ("Total millimole")/("Total volume")]` `:. "In mixture" [HCOONa]= (0.1xxV)/((V+50))` `[HCOOH]= (50xx0.05)/(V+50)` `:. pH= -log K_(a)+log ((["Salt"])/(["Acid"]))` `:. 4.0=3.80+log (((0.1xxV)//(V+50))/(2.5//(V+50)))` `:. V = 39.62 mL` |
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| 1420. |
Calculate the `pH` of `0.1M K_(3)PO_(4)`soln. The third dissociation constant of orthophoshoric acid is `1.3 xx10^(-12)`. Assume that the hydrolysis proceeds only in the first step. |
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Answer» First method: Use direct formula for the `pH` of salt of `S_(B)//W_(A)` `pH =(1)/(2)(pK_(w) +pK_(a) + logC)` `= (1)/(2) (14 +11.86 -1) =12.43` Second method: `K_(3)PO_(4) +H_(2)O hArr K_(2)HPO_(4) +KOH` or `PO_(4)^(3-) +H_(2)O hArr HPO_(4)^(2-) + overset(Θ)OH` Since hydrlysis proceeds only in I step. `:. [overset(Θ)OH] = C.h =C sqrt(((K_(w))/(K_(a).C)))=sqrt(((K_(w).C)/(K_(a))))` `K_(a)` is III dissociation constant of acid `H_(3)PO_(4)` `H_(3)PO hArr H^(o+) +H_(2)PO_(4)^(-1)...K_(1)` `H_(2)PO_(4)^(Θ) hArr H^(o+) + HPO_(4)^(2-) ...K_(2)` `HPO_(4)^(2-)hArr H^(o+) +PO_(4)^(3-) K_(3) = K_(a) = 1.3 xx 10^(-12)` `[overset(Θ)OH] = ((10^(-14)xx0.1)/(1.3xx10^(-12)))^(1//2) = 8.7 xx10^(-2)M` `:. pOH = 1.5634, pH = 12.4366` |
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| 1421. |
If pure water has `pK_(w)=13.36` at `50^(@)C`, the pH pure water will beA. `7.0`B. `7.13`C. `6.0`D. `6.63` |
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Answer» Correct Answer - D `Ph =1//2pK_(w)` (for pure water ) `pH=(1)/(2) xx 13.26 =6.63` |
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| 1422. |
At `900^(@)C`, `pK_(w)` is 13. At this temperature an aqueous solution with `pH =7` will beA. acidicB. basicC. neutralD. Unpredictable |
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Answer» Correct Answer - b |
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| 1423. |
At a certain temperature the value of `pK_(w)` is `13.4` and the measured `pH` of soln is `7`. The solution isA. acidicB. basicC. neutralD. unpredictable |
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Answer» Correct Answer - B At this temperature pH of neutral solution will be `13.4 //2` i.e., 6.7 . Thus, pH greater than 6.7 at this temperature refers to basic solution. |
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| 1424. |
The ionization constant of formic acid is `1.8xx10^(-4)`. Around what pH will its mixture with sodium formed give buffer solution of higher capacity. Calculate the ratio of sodium formate and formic acid in a buffer of `pH 4.25`. |
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Answer» For acidic capacity is `(dpH)/(dN_("acid or base"))` Thus highest buffer capacity of this is `(dpH)/(dN_("acid"))`. This will be maximum when pH is near to `pK_(a)`. Also the best results are obtained by buffer when `((["Salt"])/(["Acid"]))=10` or `(1)/(10)` i.e., within the range `pK_(a)+-1` Also `pH= -log K_(a)+log (["Salt"])/(["Acid"])` `4.25= -1.8xx10^(-4)+ log((["Salt"])/(["Acid"]))` or `4.25=3.74+log ((["Salt"])/(["Acid"]))` `:. ((["Salt"])/(["Acid"]))=3.24` |
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| 1425. |
At `900^(@)C`, `pK_(w)` is 13. At this temperature an aqueous solution with `pH =7` will beA. AcidB. BasicC. NeutralD. None of these |
| Answer» Correct Answer - 2 | |
| 1426. |
How many mole of `NH_(4)CI` must be added to one liltre of `1.0M NH_(4)OH` to have a buffer of `pH=9`? `(K_(NH_(4)OH)=1.8xx10^(-5))` |
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Answer» For basic buffer solutions `pOH= -log K_(b)+log ((["Salt"])/(["Base"]))` Let a mole of `NH_(4)CI` are added to `(1.0xx1)` mole of `NH_(4)OH` 1 litre. Given `pH= 9 , :. pOH =5` `:. 5= -log 1.8xx10^(-5)+log((a)/(1.0xx1))` or `a= 1.8` |
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| 1427. |
pH of 0.1 M NaA solution is : Given : `(K_b)_(A^-)=10^(-9)`A. 5B. 11C. 9D. 8 |
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Answer» Correct Answer - c |
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| 1428. |
pH when solution containing HA `(K_(a)=10^(-6))` and NaA show maximum buffer action will be :A. 6B. 7C. ` lt 6`D. `gt 7` |
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Answer» Correct Answer - a |
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| 1429. |
What is the pH of a 0.20 M HA solution `(K_a=1.0xx10^(-6))` that contains 0.40 M NaA ?A. 3.15B. 3.35C. `5.70`D. `6.30` |
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Answer» Correct Answer - d |
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| 1430. |
10 mL of 0.2 M HA is titrated with 0.2 M NaOH solution. Calculate change in pH between 50% of equivalence point to equivalence point. `[K_(a)` of HA `=10^(-5)]`A. 3.5B. 2C. 4D. 3 |
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Answer» Correct Answer - c |
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| 1431. |
20 mL of a weak monobasic acid (HA) requires 20 mL 0.2 M NaOH for complete titration. If pH of solution upon addition of 10 mL of this alkali to 25 mL of the above solution of HA is 5.8, then `pK_(a)` of the weak acid is :A. `6.1`B. `5.8`C. `5.98`D. `5.58` |
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Answer» Correct Answer - c |
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| 1432. |
What should be the concentration of NaA, if its 50 mL solution of `0.10MNH_(3)` and `0.10MNH_(4)CI` without changing the pH by more than `1.0` unit ? Assume no change in volume. `(K_(a) for HA= 1.0xx10^(-5))` |
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Answer» Let a be the concentration of NaA. After mixing, `M_(NaA)=(axx50)/(100)=(a)/(2)` and `M_(HA)=(0.2xx50)/(100)=0.1` Now `pH= -log K_(a)+log((["Salt"])/(["Acid"]))` `4= -1.0xx10^(-5)+log ((a//2)/(0.1))` `a=0.2M` |
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| 1433. |
How many moles of NaOH can be added to `1.0L` solution of `0.10MNH_(3)` and `0.10MN_(4)CI` without changing in volume. `(K_(b)for NH_(3)=1.8xx10^(-5))` |
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Answer» For buffer of `NH_(3)` and `NH_(4)CI`, `pOH= -log 1.8xx10^(-5)+log((1)/(1))` `pOH= 4.75` Original `pOH=4.75`, on addition of NaOH, it will change only 1 unit (decrease) i.e., new `pOH= 3.75` or pH will increase from `9.25` to `10.25`. `OH^(-)NH_(4)^(+)rarrNH_(3)+H_(2)O` `:. 3.75= -log 1.8xx10^(-5)+log (([NH_(4)^(+)])/(NH_(3)])` `([NH_(4)^(+)])/([NH_(3)])=0.1` Thus NaOH addition must not change the concentration ration by more than `0.1` i.e., `[NH_(4)^(+)]=0.1[NH_(3)]` ......(1) Initially `[NH_(4)^(+)]+[NH_(3)]=0.1+0.1=0.2` ......(2) By eqs. (1) and (2), `0.1xx[NH_(3)]+[NH_(3)]=0.2` `:. [NH_(3)]=0.182` `:. [NH_(4)^(+)]=0.018` Thus new solution must have `[NH_(4)^(+)]=0.018` in place of `0.10`. Therefore additio of `0.1-0.82` mole of NaOH be made. |
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| 1434. |
A solution of weak acid HA was titrated with base NaOH. The equivalent point was reached when 40 mL. Of 0.1 M NaOH has been added. Now 20 mL of 0.1 M HCl were added to titrated solution, the pH was found to be 5.0 What will be the pH of the solution obtained by mixing 20 mL of 0.2 M NaOH and 20 mL of 0.2 M HA?A. `7`B. `9`C. `10`D. `11` |
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Answer» Correct Answer - B `HA + NaOH rarr NaA + H_(2)O` milli moles of salt `NaA` or `A^(-) = 40 xx 0.1 = 4` `{:("Now",A^(-)+,H^(+)rarrHA),("Initial milli moles",4,2),("Final milli moles",2,-2):}` Acidic buffer solution is formed and `[A^(-)] = [HA]` `pH = pK_(a) + "log" ([A^(-)])/([HA]) rArr pK_(a) = 5` Now `HA + NaOH rarr NaA + H_(2)O`, hydrolysis of `A^(-)` will takes place `[NaA] = ("milli moles of acid")/("total volume") ` `= (20 xx 0.2)/(20 + 20) = 0.1` `pH = 1/2 (pK_(w) + pK_(a) + "log" C) = (1)/(2) [14+5-1] = 9` |
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| 1435. |
A weak acid `(HA)` is titrated with `N//100 NaOH`. What will be the `pH` when `50%` of titration is completed. Given `K_(a)=1^(-4)` & concentration of `HA=0.1M`A. 4B. 8C. 6.9D. 10 |
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Answer» Correct Answer - 1 `{:(HA,+,NaOH,rarr,NaA,+,H_(2)O),("Weak acid",,"Strong base",,"Salt of weak acid & strong base.",,):}` Upon `50%` neutralization the amount of weak acid left and that of salt formed will be same and the system will act as an acidic buffer. `{:(,HA,+,NaOH,rarr,NaA,+,H_(2)O),("Upon 50% Neutralization",a//2,,,,a//2,,):}` `pH=PK_(a)+log.([Sal t])/([Aci d])` `=4+log.(a//2)/(a//2)=4 " "Ans.(1)` |
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| 1436. |
The `pH` of an acidic buffer mixture is:A. ` gt 7`B. ` lt 7`C. `=7`D. depends upon `K_(a)` of acid |
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Answer» Correct Answer - d |
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| 1437. |
The `pH` of an acidic buffer mixture is:A. `gt7`B. `lt7`C. `=7`D. depends upon `K_(a)` of acid |
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Answer» Correct Answer - D `pH=pK_(a)+log([Sal t])/([Acid])` |
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| 1438. |
Which one of following will have the largest `pH`?A. Solution containing `1 xx 10^(-2)mol of K_(2)SO_(4)L^(-1)`.B. Pure water.C. Solution containing `1.0 xx 10^(-2)mol of HCI L^(-1)`.D. Solution containing `1 xx 10^(2)mol of NH_(4)OH L^(-1)`. |
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Answer» Correct Answer - D i. `K_(2)SO_(4)` is salt of `S_(A)//S_(B)` and do not hydrolyse so `pH = 7` ii. pH of pure `H_(2)O = 7` iii. pH of `10^(-2)` mols of HCI per litre = 2 iv. `pH` of weal base `NH_(4)OH` will be more than 7. |
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| 1439. |
`0.002` moles of an acid is added to a litre of buffer solution, decreases the pH of the buffer by `0.02`. Then the buffer capacity isA. `0.2`B. `0.1`C. `0.6`D. `0.4` |
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Answer» Correct Answer - B Buffer capacity `= ("no. of moles of an acid added per lit solution")/(P^(H) "change")` |
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| 1440. |
Which may be added to one litre of water to act a buffer?A. One mole of `CH_(3)COOH` and one mole of `HCl`B. One mole of `NH(4)OH` and one mole of `NaOH`.C. One mole of `NH_(4)Cl` and one mole of `HCl`D. One mole of `CH_(3)COOH`and `0.5`mole of `NaOH`. |
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Answer» `HC_(2)H_(3)O_(2)+NaOH+C_(2)H_(2)O_(2)Na+H_(2)O` `{:(n_(i),1,0.5,0,0),(n_(f),0.5,0,0.5,0.5):}` This solution contain weak acid & its salt with strong base and thus acts as buffer. |
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| 1441. |
Which may be added to one litre of water to act a buffer?A. One mole of `CH_(3) COOH` and one mole of HCIB. One mole of `NH_(4)OH` and one mole of NaOHC. One mole of `CH_(3)COOH` and 0.5 mole of NaOHD. One mole of `CH_(3) COOH ` and 0.5 mole of NaOH |
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Answer» Correct Answer - d |
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| 1442. |
Which may be added to one litre of water to act as a bufferA. One mole of `HC_(2)H_(3)O_(2)` and 0.5 mole of NaOHB. One mole of `NH_(4)Cl` and one mole of HClC. One mole of `NH_(4)OH` and one mole of NaOHD. One mole of `HC_(2)H_(3)O_(2)` and one mole of HCl |
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Answer» Correct Answer - A One mole oxalic acid & 0.5 mole of NaOH will make. |
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| 1443. |
Which may be added to one litre of water to act a buffer?A. One mole of `HC_(2)H_(3)O_(2)` and one mole of HCIB. One mole of `NH_(4)OH` and one mole of NaOHC. One mole of `NH_(4)CI` and mole of HCID. One mole of `HC_(2)H_(3)O_(2)` and `0.5` mole of NaOH |
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Answer» Correct Answer - D `{:(HC_(2)H_(3)O_(2)+,NaOHrarr,C_(2)H_(3)O_(2)Na+,H_(2)O),(1,0.5,0,0),(0.5,0,0.5,0.5):}` This solution contains, `underset(0.5 Mol e)("weak acid")+underset(0.5 Mol e)("its salt")` and thus acts as buffer. |
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| 1444. |
Which of the following will have the largest `pH`?A. `M//10 HCI`B. `M//100 HCI`C. `M//10NaOH`D. `M//100NaOH` |
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Answer» Correct Answer - C Largest `pH`, will be of the most basic solution so `(M)/(10) NaOH`, will have largest `pH`. |
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| 1445. |
Isoelectric point is defined as the pH at which:A. an amino acid becomes acidicB. an amino acid becomes basicC. zwitterio has positive chargeD. zwitterion has zero charge |
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Answer» Correct Answer - D Isoelectric point is the condition, where zwitterions or sol particles do not move under the influence of electric field, i.e., they lose their charge. |
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| 1446. |
Conjugate base of `overset(Theta)OH` isA. `H_(2)O`B. `H_(3)O^(o+)`C. `H^(o+)`D. `O^(2-)` |
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Answer» Correct Answer - D Conjugate base `overset(Theta)OH is O^(2-)`, `overset(Theta)OH rarr O^(2-) + H^(o+)` |
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| 1447. |
Acidic solution is defined as a solution whose `[H^(o+)] gt [overset(Theta)OH]`. Base solution has `[overset(Theta)OH] gt [H^(o+)]`. During acid-base titrations, `pH` of the mixture will change depending on the amount base added. This variation is shown in the form of graph by making plot as titration curves `100mL` of `1.0 M H_(3)A (K_(a_(1)) = 10^(-3), K_(a_(2)) = 10^(-5), K_(a_(3)) = 10^(-7))` is titrated against `0.1M NaOh`. The titration curve is as follows. What is the `pH` at point `A?A. `3`B. `4`C. `5`D. `6` |
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Answer» Correct Answer - B Since `K_(a_(1))` and `K_(a_(2))` are high values therfore point `A, K_(a_(1))` and `K_(a_(2))` are considered. `H_(3)A hArr H^(o+) + H_(2)A^(Theta)` `H_(2)A^(Theta) hArr H^(o+) + HA^(2-)` `K_(a_(1)) xx K_(a_(2)) = ([H^(o+)][HA^(2-)])/([H_(2)A]) [["At equivalence point",],[[HA^(2-)]=,[H_(3)A]]]` `-log K_(a_(1)) - log K_(a_(2)) =- 2log [H^(o+)]` `pK_(a_(1)) + pK_(a_(2)) = 2pH`. `:. pH = (pK_(a_(1)) + pK_(a_(2)))/(2) = (3+5)/(2) = 4` |
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| 1448. |
Conjugate base of `overset(Theta)OH` isA. `H_(2)O`B. `H_(3)O^(+)`C. `H^(+)`D. `O^(2-)` |
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Answer» Correct Answer - D `OH^(-) rarr H^(+) + underset(("Conjugate base"))(O^(-2))` |
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| 1449. |
In qualitative analysis, cations of group II as well as group IV precipitated in the form of sulphides. Due to low value of `Ksp` of group `II` sulphides, group reagent is `H_(2)S` in presence of dil. `HCl` and due to high value of `Ksp` of group IV sulphides, group reagent is `H_(2)S` in presence of `NH_(4)OH` and `NH_(4)Cl`. In `0.1M H_(2)S` solution, `Sn^(2+), Cd^(2+)` and `Ni^(2+)` ions are present in equimolar concentration `(0.1M)`. Given : `Ka_(1)(H_(2)S) = 10^(-7), Ka_(2)(H_(2)S) = 10^(-14), K_(sp)(SnS) = 8 xx 10^(-29) , K_(sp) (CdS) = 10^(-28), K_(sp)(NiS) = 3 xx 10^(-21)` At what pH precipitate of NiS will formA. `12.76`B. `7`C. `1.24`D. `4` |
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Answer» Correct Answer - C `k_(sp) (NiS) = 3 xx 10^(-21) :. [S^(-2)] = (3 xx 10^(-21))/(0.1)` `{:(H_(2)ShArrH^(+)+HS^(-1),k_(a_(1))=10^(-7)),(HS^(-1)hArrH^(+)+S^(-2),k_(a_(2))=10^(-14)):}` `k_(a_(2)) = ([H^(+)][S^(2)])/([HS^(-1)])` `k_(a_(1)).k_(a_(2)) = 10^(-21)([H^(+)]^(2)[3xx10^(-20)])/([0.1])` `[H^(+)] =sqrt((1)/(300)):.pH = 1.2388` |
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| 1450. |
Calculate the `pH` of a solution made by mixing `0.1 M NH_(3)` and `0.1M (NH_(4))_(2)SO_(4). (pK_(b)` of `NH_(3) = 4.76)` |
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Answer» It is a basic buffer. `["Base"] = [NH_(3)] = 0.1m` `["Salt"] = [(NH_(4))_(2)SO_(4)] = 0.1 xx2 = 0.2M` `[(NH_(4))_(2) SO_(4) rarr 2overset(o+)NH_(4) +SO_(4)^(2-)]` `pOH = pK_(b) + "log"(["Salt"])/(["Base"])` `4.76 +log (0.2M)/(0.1M)` `= 4.76 +log2` `= 4.76 +0.3 = 5.03` `:. pH = 14 - 5.03 = 8.97` |
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