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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.
| 301. |
0.5 L of `0.5 M CH_(3)COOH` and 0.5 L of 0.25 M `CH_(3)COONa` are available. What maximum volume of a buffer with a pH =4.58 can be prepared using only these two resources ? `K_(a) =1.8xx10^(-5)`A. 0.85 LB. 0.65 LC. 0.45 LD. 0.95 L |
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Answer» Correct Answer - a |
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| 302. |
0.1 M HCl and `0.1 M H_(2)SO_(4)`, each of volume 2 ml are mixed and the volume is made up to 6 ml by adding 2ml of 0.01 N NaCl solution. The pH of the resulting mixture isA. 1.17B. `1.0`C. 0.3D. log 2 - log 3 |
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Answer» Correct Answer - B Millimoles of `H^(+) = 0.1 xx 2 + 0.1 xx 2 xx 2 = 0.6` Total volume in ml = 6 `pH = -log [H^(+)] = -log ((0.6)/(6)) = -log 0.1 = 1.0`. |
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| 303. |
To a `10mL` of `10^(-3) N H_(2)SO_(4)` solution water has been to make the total volume of one litre. Its `pOH` would be `:`A. 3B. 12C. 9D. 5 |
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Answer» Correct Answer - 3 `N_(1)V_(1)=N_(2)V_(2)` `10^(-3)xx10=N_(2)xx1000` `rArrN_(2)=10^(-5)` `:. pH=5` and `pOH=14-5=9` |
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| 304. |
Assertion (A): `Sb^(3+)` is not precipitated as sulphide when `H_(2)S (g)` is passed in alkaline solution. Reason (R) : `[S^(2-)]` ion in basic medium is inadequate for precipitation.A. Both A and R true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - C | |
| 305. |
The volume of `0.1 M Na_(2)SO_(4)` needed to be added to 10 mL of 0.1 M HCI to change the pH from 1.00 to 1.50 is : `(K_(2) =1.26 xx10^(-2) " for " H_(2)SO_(4))`A. 6.6 mLB. 4.4 mLC. 2.2 mLD. 1.1 mL |
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Answer» Correct Answer - a |
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| 306. |
The ratio of `pH` of solution `(1)` containing 1 mole of` CH_(3)COONa` and 1 mole of `HCl` and solution `(II)` containing 1 mole of `CH_(3)COONa` and 1 mole of acetic acid in one litre is `:` |
| Answer» Correct Answer - `1:2;` | |
| 307. |
Phenol `(C_(6)H_(5)OH,K_(a)=1.3xx10^(-10))` is a weak acid used in mouth washes and pyridine `(C_(5)H_(5)N,K_(b)=1.8xx10^(-9))` is a weak base used as a solvent. Calculate the value of `K_(n)` for neutralization of phenol by pyridine. Does the neutralization reaction proceed very far towards completion ? |
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Answer» `C_(6)H_(5)OH+C_(5)H_(5)N hArr C_(6)H_(5)O^(-)+C_(5)H_(5)NH^(+)` `K_(H)= (K_(w))/(K_(a)xxK_(b))= (10^(-14))/(1.3xx10^(-10xx1.8xx10^(-9))` `=4.273xx10^(4)` `K_(n)=(1)/(K_(H))=(1)/(4.273xx10^(4))=2.34xx10^(-5)` `K_(n)` being small and thus neutralisation does not proceed very far towards completion. |
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| 308. |
Assertion `(A ) :` A buffer solution has a capacity to resist the change in pH value on addition is small amount of acid or base to it. Reason `(R ) :` An aqueous solution of ammonium acetate can act as a buffer.A. Both A and R true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - B | |
| 309. |
Calculate the pH of a buffer solution prepared by dissolving 30g of `Na_(2)CO_(3)` in 500 mL of an aqueous solution contaning 150 mL of 1 M HCI. `(K_(a) for HCO_(3)^(-)= 5.63xx10^(-11))` |
| Answer» Correct Answer - `10.197;` | |
| 310. |
`pOH` of `0.002 MHNO_(3) `is `:`A. `11+log2`B. `11-log2`C. `-3 +log 2`D. none of these |
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Answer» Correct Answer - 1 `[H^(+)]=0.002M` `pH=log2xx10^(-3)=3-log2` `:.pOH=11+log2` |
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| 311. |
Which is a buffer solutionA. `CH_(3)COOH + CH_(3)COONa`B. `CH_(3)COOH + CH_(3)COONH_(4)`C. `CH_(3)COOH + NH_(4)Cl`D. `NaOH + NaCl` |
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Answer» Correct Answer - A Buffer solution is a mixture of weak acid and its conjugate base. |
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| 312. |
The percentage of pyridine `(C_(2)H_(5)N)` that forms pyridinium ion `(C_(2)H_(5)N^(+)H)` in a 0.10 M aqueous pyridine solution (`K_(b)` for `C_(2)H_(5)N = 1.7 xx 10^(-9)`) isA. `1.6%`B. `0.0060%`C. `0.013%`D. `0.77%` |
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Answer» Correct Answer - C `C_(5)H_(5)N + H_(2)O = C_(5)H_(5)N^(+)H + OH^(-)` 0.1 `alpha = sqrt((K_(b))/(c)) = sqrt((1.7 xx 10^(-9))/(0.1)) = sqrt(1.7 xx 10^(-8)) = 1.3 xx 10^(-4)` `% prop = 1.3 xx 10^(-4) xx 100` `= 1.3 xx 10^(-2) = 0.013`. |
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| 313. |
A `0.25 M` solution of pyridinium chloride `C_(5)H_(5)NH^(+)Cl^(-)` was found to have a `pH` of `2.75` What is `K_(b)` for pyridine, `C_(5)H_(5)N`? |
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Answer» `C_(5)H_(5)NH^(+)Cl^(-)`" " It is salt of `SAWB` Pyridinium ion on hydrolusis will produce `H_(3)O^(+)` `:.[H_(3)O^(+)]=antilog(-2.75)=Ch=0.25h` `h` comes negligible `(lt0.1)` `pH=7-(1)/(2)pK_(b)-(1)/(2)logCrArr pH=7-(1)/(2)pK_(b)-(1)/(2)log0.25` `rArr pK_(b)=9.1` `K_(b)=Antilog (-9.1)=8xx10^(-10)` |
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| 314. |
`0.15` mole of pyridinium chloride has been added into `500 cm^(3)` of `0.2M` pyridine solution. Calculate pH and hydroxyl ion contration in the resulting solution, assuming no change in volume. `(K_(b)"for pyridine" = 1.5xx10^(-9)M)` |
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Answer» `["Pyridinium chlodide"]= (0.15//500)xx1000=0.3M` `["Pyridine"]= 0.2M` :. A mixture of pyridine and its salt pyridinium chloride forms a basic a buffer and therefore, `pOH= -log K_(b)+log ["Salt"]//["Base"]` or `pOH= -log 1.5xx10^(-9)+log(0.30//0.20)` `= -log 1.5+9 log 10+log 1.5=9` `:. [OH^(-)]=10^(-9)` and `[H^(+)]=10^(-5)` So `pH=5` |
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| 315. |
MY and `NY_(3)` two nearly insoluble salts, have the same `K_(sp)` values of `6.2xx10^(-13)` at room temperature. Which statement would be true in rearged to MY and `NY_(3)` ?A. The molar solubility of MY in water is less than that of `NY_(3)`.B. The salts MY and `NY_(3)` are more soluble in 0.5 M KY than in pure waterC. The addition of the salt of KY to solution of MY and `NY_(3)` will have no effect on their solubilitiesD. The molar solubilities of MU and `NY_(3)` in water are identical. |
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Answer» Correct Answer - A For MY, `underset(0)(MY)hArr underset(s)(M^(+))+underset(s)(Y^(-))` where, s = solubility and `K_(sp)` = solubility product. `therefore K_(sp)=[M^(+)][Y^(-)]` `K_(sp)=s^(2)` `s=sqrt(K_(sp))=sqrt(6.2xx10^(-13))=7.874xx10^(-7)` Similarly, for `NY_(3)`, `underset(0)(NY_(3))hArr underset(s)(N^(+))+underset(3s)(3Y^(-))` `therefore K_(sp)=[N^(+)][Y^(-)]^(3)=sxx(3s)^(3)` `K_(sp)=27s^(4)` `therefore s= root(4)((K_(sp))/(27))=root(4)((6.2xx10^(-13))/(27))=3.89xx10^(-4)` Therefore, molar solubility of MY in water is less than that of `NY_(3)`. |
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| 316. |
Determine degree of dissociation of `0.05M NH_(3)` at `25^(@)C` in a solution of `pH = 11`. |
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Answer» Correct Answer - 2 `{:(NH_(4)OHhArr,NH_(4)^(+)+,OH^(-)),(1,0,0),((1-alpha),alpha,alpha):}` Given `pH =11` `:. [H^(+)]=10^(-11)` `:. [OH^(-)]=10^(-3)=c.alpha` Since `c=0.05` `alpha= (10^(-3))/(c )=(10^(-3))/(0.05)=0.02=2%` |
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| 317. |
The solubility of calomel in water at `25^(@)C` is x mole `//` litre. Its solubility product isA. `4x^(3)`B. `12x^(3)`C. `108x^(5)`D. `x^(2)` |
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Answer» Correct Answer - A `Hg_(2)Cl_(2) hArr underset(x) (Hg_(2)^(+2))+underset(2x)(2Cl^(-))` `K_(sp)=[Hg_(2)^(+2)][Cl^(-)]^(2)=(x)(2x)^(2)=4x^(3)` |
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| 318. |
The addition of HCl will not supress the ionisation ofA. acetic acidB. `H_(2)S`C. benzoic acidD. sulphuric acid. |
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Answer» Correct Answer - D Because `H_(2)SO_(4)` is a strong acid. |
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| 319. |
Which of the following will produce a buffer sollution when mixed in equal volumes ?A. `0.1 " mol dm"^(-3) NH_(4)OH " and " 0.1 " mol dm"^(-3) HCI`B. `0.05 " mol dm" ^(-3) NH_(4)OH " and " 0.1 " mol dm"^(-3) HCI`C. `0.1 " mol dm"^(-3) NH_(4)OH " and " 0.05 " mol dm"^(-3) HCI`D. `0.1 " mol dm " ^(-3) CH_(4)COONa " and " 0.1 " mol dm" ^(-3) NaOH` |
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Answer» Correct Answer - c |
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| 320. |
When an acid or alkali is mixed with buffer solution, then pH of buffer solutionA. Not changesB. Changes slightlyC. IncreasesD. Decreases |
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Answer» Correct Answer - A Because the pH of buffer are not changed. |
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| 321. |
If concentration of two weak acids are `C_(1)` and `C_(2) "mol"//L` and degree of ionization are `alpha_(1)` and `alpha_(2)` respectively then their relative strength can be compared by :A. `([H^(+)]_(1))/([H^(+)]_(2))`B. `(alpha_(1))/(alpha_(2))`C. `(C_(1)alpha_(1))/(C_(2)alpha_(2))`D. `(K_(a_(1))C_(1))/(k_(a_(1))C_(2))` |
| Answer» Correct Answer - A::C | |
| 322. |
pH of a solution can be expressed asA. `-log_(e)(H^(+))`B. `-log_(10)(H^(+))`C. `log_(e)(H^(+))`D. `log_(10)(H^(+))` |
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Answer» Correct Answer - B `pH = -log [H^(+)]`. |
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| 323. |
Correct statements about the percentage ionization of weak acids in water include which of the following ? (P)The percentage ionization increases as the ionization constant of the acid becomes larger. (Q)the percentage ionization increases as the concentration of the acid becomes smaller.A. P onlyB. Q onlyC. Both P and QD. Neither P nor Q |
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Answer» Correct Answer - c |
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| 324. |
The ionization constant of a certain weak acid is `10^(-4)`. What should be the [salt] to [acid] ratio if we have to prepare a buffer with pH = 5 using this acid and one of the saltsA. `1:10`B. `10:1`C. `5:4`D. `4:5` |
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Answer» Correct Answer - B `pH = -log K_(b) + log.(["salt"])/(["acid"])` `5 = - log 10^(-4) + log.(["salt"])/(["acid"])` `log.(["salt"])/(["base"]) = 1` `(["salt"])/(["base"]) = "antilog 1" = 10 : 1`. |
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| 325. |
The percentage ionization of a weak base is given byA. `(sqrt((K_(a))/c))xx 100`B. `((1)/(1+10^(pK_(b)-pOH))) xx 100`C. `(sqrt((K_(b))/c))xx 100`D. `(sqrt((K_(w))/(c xx K_(a) "of conjugate acid")))100` |
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Answer» Correct Answer - C::D Let weak base be `= BOH hArr B^(+) + OH^(-)` `{:("Initial conc.",c,0,0,),("Final conc.",c-c alpha,c alpha,c alpha,):}` Let bolume of container be 1 litre `rArr` % dissociation `= (c alpha)/(c) xx 100 = 100 alpha` Also `(c alpha^(2))/(1-alpha) = K_(b)` As `alpha` is negligible `rArr K_(b) = c alpha^(2) , alpha = sqrt((K_(b))/(c))` `rArr` % ionization `= 100 alpha = (sqrt((K_(b))/(c))) 100` `rArr 100alpha = (sqrt((K_(b))/(c))) 100 = (sqrt((K_(w))/(K_(a)c))) 100` `[ because K_(b) = (K_(w))/(K_(a))]` (where `c alpha = [OH^(-)]`) `= log_(10) c alpha = pOH = -log_(10) (K_(b)c)^(1//2) rArr pOH = -(1)/(2)log K_(b)c` `pOH = -(1)/(2)log K_(b)c = -log c alpha` `rArr pOH = (1)/(2)[pK_(b) - log c] = -log c alpha` `2pOH = pK_(b) - logc rArr = pK_(b) - 2pOH` `rArr c = 10^(pK_(b) - 2pOH)` Hence choices (c) and are correct while (a) and (b) are not correct. |
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| 326. |
Which of the following solution (s) is/are correctA. In contains weak acid and its conjugate baseB. In contains weak base and its conjugate acid.C. It shows change in pH on adding small amount of acid or baseD. All of these |
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Answer» Correct Answer - A::B (a) A buffer solution is a solution which contains weak acid and its conjugate base. It is acidic buffer. (b) Basic buffer contains weak base and its conjugate acid. (c) Is wrong because it does not show change in pH on adding small amount of acid or base. (d) Is wrong, all of the above statements are not correct. |
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| 327. |
The pH of basic buffer mixtures is given by pH `= pK_(a) + log.(["Base"])/(["Salt"])`, whereas pH of acidic buffer mixtures is given by : pH `= pK_(1) + log.(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio `(["Base"])/(["Salt"])` for `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. The volume of 0.2 M NaOH needed to prepare a buffer of pH 4.74 with 50 ml of 0.2 M acetic acid (`pK_(b)` of `CH_(3)COO^(-) = 9.26`) isA. 50 mLB. 25 mLC. 20 mLD. 10 mL |
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Answer» Correct Answer - B Let V mL of NaOH be needed to give `CH_(3)COONa`. `{:(NaOH+,CH_(3)COO,rarr,CH_(3)COONa+,H_(2)O),(0.2 xx V,50 xx 0.2,," "0," "0),(-,[10-0.2V],," "0.2V,0.2 V):}` `:. pH = pK_(a) + log.(["Salt"])/(["Acid"]) = pK_(w) - pK_(b) + log.(["Salt"])/(["Acid"])` `= 14 - 9.26 + g.(["Salt"])/(["Acid"])` `= 14-9.26 + log.([(0.2V)/(50+V)])/([(10-0.2 V)/(50+V)])` `4.74 = 4.74 + log [(0.2V)/(10 - 0.2 V)] :. V = (10)/(0.4) = 25 mL`.` |
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| 328. |
There is a salt of weak acid and weak base the `K_(a)` of the acid is greater than the `K_(b)` of the base henceA. Degree of hydrolysis is independent of the concentration of the salt solutionB. Anionic hydrolysis will be greater than cationic hydrolysisC. Cationic hydrolysis is greater than anionic hydrolysisD. `h = sqrt((K_(w))/(K_(a)K_(b)))` |
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Answer» Correct Answer - A::C::D Foa a salt of weak acid and weak base `K_(h) = (K_(w))/(K_(a)K_(b)) = ([HA][BOH])/([A^(-)][B^(+)]` and `h = sqrt((K_(w))/(K_(a)K_(b)))` Hence degree of hydrolysis is independent of the concentration of the salt taken. `{:("Further",A^(-)+H_(2)O hArr HA+OH^(-),,....(i)),(,B^(+)+H_(2)O hArr BOH + H^(+),,...(ii)):}` Since `K_(a)` of HA is more than `K_(b)` of BOH so the extent of backward reaction in (i) is greater than the extent of of hydrolysis of `B^(+)` Hence choices (a), (c) and (d) are correct while (b) is incorrect. |
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| 329. |
The pH of basic buffer mixtures is given by pH `= pK_(a) + log.(["Base"])/(["Salt"])`, whereas pH of acidic buffer mixtures is given by : pH `= pK_(1) + log.(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio `(["Base"])/(["Salt"])` for `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. The ratio of pH of solution (I) containing 1 mole of `CH_(3)COONa` and 1 mole of acetic acid in one litre isA. `1:2`B. `2:1`C. `1:3`D. `3:1` |
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Answer» Correct Answer - A `{:(CH_(3)COONa+,HCl rarr,CH_(3)COOH+,NaCl,),(" "1," "1," "0," "0,),(" "0," "0," "1," "1,):}` `:. [CH_(3)COOH] = (1)/(2) =1`. `:. [H^(+)] = C alpha = C sqrt((K_(a))/(C)) = sqrt(K_(a).C) = sqrt(K_(a))` or `pH_(1) = -(1)/(2) log K_(a) = (1)/(2) pK_(a)` `{:(CH_(3)COOH+,CH_(3)COONa),(" "1," "1):}` `:. pH = pK_(a) + log.(1)/(1) rArr pH_(2) = pK_(a) :. (pH_(1))/(pH_(2)) = (1)/(2)`. |
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| 330. |
Statement 1 : `H_(2)SO_(4)`, HCl and `HNO_(3)` are all equally strong in water but not equally strong in acetic acid. Statement 2 : `H_(2)O` gives `H^(+)` as well as `OH^(-)` ions, but `CH_(3)COOH` gives only `H^(+)` and no `OH^(-)` ions.A. Statement 1 is true, statement 2 is true , statement 2 is correct explanation for statement 1B. Statement 1 is true, statement 2 is true , statement 2 is not a correct explanation for statement 1C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - B `H_(2)SO_(4), HCl` and `HNO_(3)` dissociate to different extent in acetic acid because acetic acid is a poor proton acceptor and hence acts as a differentiating solvent. |
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| 331. |
If `CH_(3)COOH(K_(a)=10^(-5))` reacts with NaOH at 298K, then find out the value of the maximum rate constant of the reverse reaction at 298 K at the end point of the reaction. Given that the rate constant of the forward reaction is `10^(-11)mol^(-1) L sec^(-1)` at 298 K. Also calculate Arrhenius parameter for backward reaction if `DeltaH_(298K)=44kcal` and `E_(a(f))= 94kcal`. |
| Answer» Correct Answer - `10^(-20),2.71xx10^(+16);` | |
| 332. |
Statement 1 : The molar concentrations of proton aceeptor and proton donor become equal at the midpoint of titration of a weak acid. Statement 2 : The `pK_(a)` of a weak acid becomes equal to pH of the solution at the midpoint of its titration.A. Statement 1 is true, statement 2 is true , statement 2 is correct explanation for statement 3B. Statement 1 is true, statement 2 is true , statement 2 is not a correct explanation for statement 3C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - A According to Henderson-Hasselbalch equation `pH = pK_(a) + log.(["proton acceptor"])/(["proton donor"])` At midpoint of titration of weak acid `[HA] = [A^(-)]` and hence `pH = pK_(a) + log 1.0 = pK_(a)` |
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| 333. |
The `K_(w)` of water at two temperature `25^(@)C` and `50^(@)C` are `1.08xx10^(-14), 5.474xx10^(-14)` respectively. Assuming `DeltaH` of any reaction is neutralisation of a strong acid with strong base. |
| Answer» Correct Answer - `-12.5xx10^(3)cal;` | |
| 334. |
Assertion (A): `pK_(a)` of a weak acid become equal of the `pH` of the solution at the mid-point of titration. Reason (R) : The molar concentration of the proton donor an proton acceptor beomes equal at the mid-point.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
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Answer» Correct Answer - B `(A)` is correct. At midpoint of rtitration of weak acid, `["salt"] = ["Acid"]` and therefore `pH = pK_(a)` `(R)` is correct but not the reason of `(A)`. |
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| 335. |
The `pH` of pure water at `25^(@)C` and `35^(@)C` are `7` and `6`, respectively. Calculate the heat of formation of water from `H^(o+)` and `overset(Θ)OH`. |
| Answer» Correct Answer - `84.551 kcal//mol;` | |
| 336. |
The `pH` of pure water at `25^(@)C` and `35^(@)C` are `7` and `6`, respectively. Calculate the heat of formation of water from `H^(o+)` and `overset(Θ)OH`.A. `84.55kcal mol^(-)`B. `-84.55kcal mol^(-1)`C. `74.55kcal mol^(-1)`D. `-74.55kcal mol^(-1)` |
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Answer» Correct Answer - B At `25^(@)C, [H^(+)]=10^(-7)` `:. K_(w)=10^(-14)` At `35^(@)C, [H^(+)]=10^(-6)` `:. K_(w)=10^(-12)` Now using, `2.303log (K_(w_(2))/(K_(w_(1))))=(Delta H)/(R ) [(T_(2)-T_(1))/(T_(1)xxT_(2))]` `2.303log (10^(-12)/(10^(-14)))=(Delta H)/(2)[(10)/(298xx308)]` `:. Delta H=84551.4cal//mol=84.551 kcal//mol` Thus, `H_(2)OhArr H^(+)+OH^(-)`, `Delta H=84.551 kcal//mol` `:. H^(+)+OH^(-)hArr H_(2)O`, `Delta H= -84.551 kcal//mol` |
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| 337. |
The pH of a solution obtained by mixing 10 mL of `0.1MHCI` and 40mL of `0.2MH_(2)SO_(4)` is:A. `1.4865`B. `0.4865`C. `0.4685`D. 3 |
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Answer» Correct Answer - C Milli-equivalent of `H^(+)` from `HCI=10xx0.1=1` Milli-equivalent of `H^(+)` form `H_(2)SO_(4)` `=40xx0.2xx2=16` Total Meq. Of `H^(+)` in solution `= 1+16=17` `:. [H^(+)]= (Meq.)/(V_(i nmL))=(17)/(50)=3.4xx10^(-1)` `pH= -log[H^(+)= -log 0.34=0.4685` |
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| 338. |
Assertion (A): Due to common ion effect, the solubility of `HgI_(2)` is expected to be less in an aqueous solution of `KI` than in water. But `HgI_(2)` dissolves in an aqueous solution of `KI` to form a clear solution. Reason (R) : `I^(Theta)` ions is highly polarisable.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
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Answer» Correct Answer - B `HgI_(2) + KI rarr K_(2) [HgI_(4)] (A)` is true. `(R)` is also true since `I^(Theta)` ion is large sized and therefore is highly polarisable. But `(R)` is not the correct explanation of `(A)`, |
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| 339. |
At `-50^(@)C` liquid `NH_(3)` has ionic product is `10^(-30)` .How many amide `(NH_(2)^(-))` ions are present per mm`.^(3)` in pure liqudi `NH_(3)`? (Take `N_(A)=6xx10^(23)`) |
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Answer» `K=[NH_(4)""^(+)][NH_(2)""^(=)]=10^(-30)` `[NH_(2)^(-)]=[NH_(4)^(+)]=10^(-15)M " "( :.2NH_(3)hArr NH_(4)^(+)+NH_(2)""^(-))` No. of `NH_(2)^(-)` ions `NH_(2)^(-)=((10^(-15)mol e)/(L))((1L)/(10^(6)mm^(3)))((6xx10^(23)ions)/(mol e))=600ions//mm^(3)` |
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| 340. |
What is the ionisation constant of `HOCl` if `K_(b)` of `Ocl^(-)=4xx10^(-10)` ?Also find its `pK_(a)`. |
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Answer» `K_(a)=(K_(w))/(K_(b))=(10^(-14))/(4xx10^(-10))=2.5xx10^(-5)` `pK_(a)=5-log2.5=4.6` |
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| 341. |
Calculate the `pH` of `0.08` solution of `HOCI` (hydrochlorous acid). The ionisation constant of the acid is `2.5 xx 10^(-5)`. Determine the percent dissociation of `HOCI`. |
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Answer» Using direct formula, `K_(a) = 2.5 xx 10^(-5) = 25 xx 10^(-6) = (5)^(2) xx 10^(-6)` `pK_(a) =- log [(5)^(2) xx 10^(-6)] =- 2log 5+ 6` `=- 2xx 0.7 +6 = 4.6` a. `pH = (1)/(2) (pK_(a) - log C) = (1)/(2) (4.6 - log 0.08)` `= (1)/(2) (4.6 - log (2)^(3) xx 10^(-2))` `= (1)/(2) [4.6 - 3xx0.3 +2]` `= 2.85` `:. pH = 2.85` b. `alpha = sqrt((K_(a))/(c)) = sqrt((25 xx 10^(-6))/(0.08)) = sqrt((25 xx10^(-6)xx100)/(8))` `= sqrt(25xx10^(-4))/(2sqrt(2))` `= (5xx10^(-2))/(2xx1.414) = 0.017` `= 1.7 xx 10^(-3)M` `HOCl(aq) + H_(2)O(l) hArr H_(3)O^(o+)(aq) + ClO^(ɵ)(aq)` `{:("Initial conc",rArr0.08,-,0,0),("Conc at eq",rArr0.08(1-alpha),-,Calpha,Calpha),(,~~0.08(1-alpha~~1),,,):}` `:.[HOCI] = 0.08M` ltbRgt `[H_(3)O^(o+)] = [CI^(Θ)] = Calpha = 0.08 xx 1.7 xx 10^(-3)` `= 1.41 xx 10^(-3)M` c. % dissociation `= ([HOCI]_("dissociated"))/([HOCI]_("undissociated")) xx 100` `= (1.41 xx 10^(-3))/(0.08) xx 100 = 1.76%` |
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| 342. |
HClO is a weak acid. The concentration of `H^(+)` ions in 0.1 M solution of HClO `(K_(a) = 5 xx 10^(-8))` will be equal toA. `7.07 xx 10^(-5) m`B. `5 xx 10^(-9) m`C. `5 xx 10^(-7) m`D. `7 xx 10^(-4) m` |
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Answer» Correct Answer - D `[H^(+)]^(2) = C.alpha = 0.1 xx 5 xx 10^(-8)` `H^(+) = sqrt(5 xx 10^(-9)) = 7.07 xx 10^(-5) M`. |
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| 343. |
An aqueous solution of aluminium sulphate would showA. AcidicB. NeutralC. BasicD. Both acidic and basic reaction. |
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Answer» Correct Answer - A `AI_(2)(SO_(4))_(3)` is a salt of `W_(B)//S_(A) [AI(OH)_(3)+H_(2)SO_(4)]`. It hydrolyses and given acidic solution. |
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| 344. |
Which is strongest Lewis baseA. `SbH_(3)`B. `AsH_(3)`C. `PH_(3)`D. `NH_(3)` |
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Answer» Correct Answer - D Acidity increase with increase in atomic size and basically decreaes and `NH_(3)` smallest atomic size and a lone pair of electron. |
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| 345. |
An aqueous solution of aluminium sulphate would showA. An acetic reactionB. A neutral reactionC. A basic reactionD. Both acidic and basic reaction |
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Answer» Correct Answer - A `Al_(2)(SO_(4))_(3)` is a salt of weak base `Al_(2)(OH)_(3)` and strong acid `H_(2)SO_(4)`. |
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| 346. |
The acid that results when a base accepts a proton is calledA. Conjugate base of the acidB. Conjugate protonated baseC. Lewis baseD. Conjugate acid of the base |
| Answer» Correct Answer - D | |
| 347. |
`pH` for the solution of salt undergoing anionic hydrolysis (say `CH_(3)COONa`) is given by:A. `pH=1//2[pK_(w)+pK_(a)+logC]`B. `pH=1//2[pK_(w)+pK_(a)-logC]`C. `pH=1//2[pK_(w)+pK_(b)-logC]`D. none of these |
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Answer» Correct Answer - A `CH_(3)COO^(-)+H_(2)OhArrCH_(3)COOH+OH^(-)` `:. [OH^(-)]=C.h` `=C sqrt((K_(H))/(C))=sqrt(K_(H).C)=sqrt((K_(w))/(K_(a))C)` or `-log OH= -(1)/(2)[log K_(w)++log C-log K_(a)]` or `pOH= (1)/(2)[pK_(w)-logC-pK_(a)]` Now `pH+pOH=pK_(w)` `:. pH=(1)/(2)[pK_(w)+logC+pK_(a)]` |
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| 348. |
Which of the following is the strongest Lewis acidA. `BI_(3)`B. `B Br_(3)`C. `BCl_(3)`D. `BF_(3)` |
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Answer» Correct Answer - A Larger the size of halogen atom less is the back donation of electrons into empty 2p orbital of B. |
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| 349. |
`pH` for the solution of salt undergoing anionic hydrolysis (say `CH_(3)COONa`) is given by:A. `pH=(1)/(2)pK_(w)-(1)/(2)pK_(b)-(1)/(2)log_(c )`B. `pH=(1)/(2)pK_(w)+(1)/(2)pK_(a)-(1)/(2)pK_(b)`C. `pH=(1)/(2)pK_(w)+(1)/(2)pK_(a)+(1)/(2)log _(c )`D. None |
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Answer» Correct Answer - C See hydrolysis of salts in comprehensive review. |
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| 350. |
The solvent which neither accepts proton nor donates proton is calledA. AmphotericB. NeutralC. AproticD. Amphiprotic |
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Answer» Correct Answer - C The solvent which neither accept proton nor donates. |
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