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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.
| 1551. |
When rain is accompained by a thunderstorm, the collected rain water will have a `pH`:A. Slightly lower than that of rain water without thunderstormB. sightly higher than that when the thunderstorm is not thereC. uninfluenced by occurrence of thunderstormD. which depends on the amount of dust in air |
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Answer» Correct Answer - a |
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| 1552. |
Which of the following solutions has `pH =11` ?A. `10^(-11) M NaOHB. `10^(-3)M `HClC. `10^(-3)` M NaOHD. `10^(3)` M NaOH |
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Answer» Correct Answer - C `[H^(+)]=(10^(-14))/(10^(-3))=10^(-11)` Hence `pH =-log ][H^(+)]=11` |
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| 1553. |
Which of the followin will have the largest , pH ?A. 0.1 N HClB. 0.1 N `CH_(3)COOH`C. 0.1 N NaOHD. 0.01 N NaOH |
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Answer» Correct Answer - C A strong basic solution will have largest `pH ` |
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| 1554. |
The reverse procees of neutralisation is:A. hydrolysisB. decompositionC. dehydrationD. synthesis |
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Answer» Correct Answer - A Neutralisation is exothermic, whereas hydrolysis is endothermic. |
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| 1555. |
Solubility of salt `A_(2)B_(3)` is `1 xx 10^(-4)`, its solubility product isA. `1.08 xx 10^(20)`B. `1.08 xx 10^(18)`C. `2.6 xx 10^(-18)`D. `1.08 xx 10^(-18)` |
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Answer» Correct Answer - D `{:(A_(2)B_(3)hArr,2A^(3+)+,3B^(2-),),(,2S,3S,):}` `K_(sp) = (2S)^(2) (3S)^(2) = 108 S^(5) = 108 xx (10^(-4))^(5)` `= 108 xx 10^(-20) = 1.08 xx 10^(-20) = 1.08 xx 10^(-18)`. |
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| 1556. |
When `2mol` of `HCI` is added to `1L` of an acidic buffer, its `pH` changes from `3.4` to `2.9`. The buffer capacity of the buffer solution isA. 2B. 0C. 4D. 8 |
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Answer» Correct Answer - C Buffer capacity `=` moles of acid which changes the pH of 1L solution by unity. |
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| 1557. |
Calculate the amount of `NH_(4)C1` required to dissolve in `500mL` of water to have a `pH = 4.5, K_(b) = 2.0 xx 10^(-5)`. |
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Answer» Correct Answer - C `NH_(4)C1` is a salt of `W_(B)//S_(A). (pK_(b) =- (log 2xx 10^(-5)) =- 0.3 +5 = 4.7` `:. pH = (1)/(2) (pK_(w) - pK_(b) - logC)` `4.5 = (1)/(2) (14 - 4.7 - logC)` `9 = 9.3 - logC`, `log C = 0.3` `C = "Antilog" (0.3) ~~ 2M~~2 moles//L ~~1 mol//500 mL` Weight of `NH_(4)C1 = 53.5g.//500mL (Mw NH_(4)C1 = 53.5 g mol^(-1))`. |
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| 1558. |
A buffer solution contains `0.25M NH_(4)OH` and `0.3 NH_(4)C1`. a. Calculate the `pH` of the solution. b. How much `NaOH` should be added to `1L` of the solution to change `pH` by `0.6.K_(b) =2xx10^(-5)`. |
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Answer» Correct Answer - A::B i. Since it is basic buffer. Thus, `K_(b) = 2 xx 10^(-5)` `pK_(b) =- log (2 xx 10^(-5)) =- 0.3 +5 = 4.7` `pOH = pK_(a) + "log"(["Salt"])/(["Base"])` `= 4.7 + log ((0.3)/(0.25))` `= 4.7 + log (1.2)` `= 4.7 + log (12 xx 10^(-1))` `= 4.7 + log 2^(2) + log 3 - 1` `= 3.7 + 0.6 + 0.48 = 4.78` `pH = 14 - 4.78 = 9.22` ii. Rule `B B B:` In basic buffer (B),on addition of `S_(B)(B)`, the concentration of `W_(B)(B)` increase and that of salt decreases. On adding base, `pH` will increase and `pOH` will decrease. Let `xM NaOH` is added. `pH_("initial") = 9.22`, `pH_("new") = 9.22 + 0.6 = 9.86` `pOH_("new") = 14 - 9.86 = 4.14` `["Base"]_("new") = (0.25 +x)` `["Salt"]_("new") = (0.3 - x)` `:. pOH = pK_(b) + log ((0.3 -x)/(0.25 +x))` `4.14 = 4.74 + log ((0.3 -x)/(0.25 +x))` `4.14 - 4.74 = log ((0.3 - x)/(0.25+x))` `- 0.6 = log ((0.3 -x)/(0.25 +x)) "or log" ((0.25 +x)/(0.3-x)) = 0.6` `:. (0.25+x)/(0.3 -x) = "Antilog" (0.6) = 4` Solve for `x :` `x = 0.19` mol in `1L` of solution |
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| 1559. |
Which of the following expression is wrong?A. `[H^(o+)] = [overset(Theta)OH] = 10^(-7)` for a neutral solution at all temperatures.B. `[H^(o+)] lt sqrt(K_(w))` and `[overset(Theta)OH] gt sqrt(K_(w))` for basic solutionC. `[H^(o+)] = [overset(Theta)OH] = sqrt(K_(w))` for a neutral solutionD. `[H^(o+)] gt sqrt(K_(w))` and `[overset(Theta)OH] lt sqrt(K_(w))` for an acidic solution |
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Answer» Correct Answer - A Statement (a) is wrong since it is not valid all temperature. |
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| 1560. |
How many moles of `NH_(4)C1` should be added to `200mL` solution of `1.18 M NH_(4)OH` to have a `pH` of `9.60. K_(b)` of `NH_(4)OH= 2 xx 10^(-5)` |
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Answer» Correct Answer - A It is basic buffer. `pH = 9.6, pOH = 14 - 9.6 = 4.4` `K_(b) = 2 xx 10^(-5), pK_(b) =- log (2xx10^(-5)) =- 0.3 +5 = 4.7` `pOH = pK_(b) + "log"(["Salt"])/(["Base"])` `4.4 = 4.7 + "log"(["Salt"])/((0.18))` `- 0.3 = log["Salt"] - log (0.18) = log ["Salt"] + 0.7447` `- 0.3 - 0.7447 = log ["Salt"]` `:. log ["Salt"] =- 1.0447` `["Salt"] = "Antilog" (-1-0.447+1-1)` `= "Antilog" (bar(2).9553) = 9.55 xx 10^(-2)M`. Moles of salt `= (9.55 xx 10^(-2) xx 200)/(1000)` `= 1.91 xx 10^(-2)= 0.019` mol. |
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| 1561. |
A is an aqueous acid, B is an aqueous base. They are diluted separately, thenA. pH of A increases and pH of B decreasesB. pH of A inceases and pH of B decreases till pH in each case is 7C. pH of A and B increaseD. pH of A and B decrease |
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Answer» Correct Answer - A On adding the pH of acid A increases while pH of base B decreases. |
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| 1562. |
Select the `pK_(a)` value of the strongest acid from the followingA. `1.0`B. `3.0`C. `2.0`D. 4.5 |
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Answer» Correct Answer - A `pK_(a) lt lt` then stronger acid `pK_(a) gt gt` then weak acid `pK_(a) prop (1)/("Acidic strength")`. |
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| 1563. |
`pK_(a)` value of the strongest acid among the following isA. `3.0`B. `4.5`C. `1.0`D. `2.0` |
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Answer» Correct Answer - C Smaller the `pK_(a)` value than stronger the acid. |
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| 1564. |
Heat of neutralisation of weak acid and strong base is less than the heat of neutralisation of strong acid and strong base due toA. Energy has to be spent for the total dissociation of weak acidB. Salt of weak acid and strong base is not stableC. Incomplete dissociation of weak acidD. Incomplete neutralisation of weak acid |
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Answer» Correct Answer - A Energy has to be spent for the total dissociation of weak acid. |
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| 1565. |
Determine whether `Cd^(2+)` can be separed form `Zn^(2+)` by bubbling `H_(2)S` through a `0.3M` HCI solution that contgains `0.005M Cd^(2+)` and `0.005M Zn^(2+)` . `(K_(SP)` for CdS and ZnS are `8xx10^(-7)` and `3xx10^(-2))` respectively) |
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Answer» On bubbling `H_(2)S`, the solution becomes saturated with `H_(2)S` to give MS and thus `[H_(2)S]=0.005=[M^(2+)]` `M^(2+)+H_(2)S+2H_(2)OhArr MS_((s))+2H_(3)O_((aq.))^(+)` `Q_(f)=([H_(3)O^(+)]^(2))/([M^(2+)][H_(2)S])= ((0.3)^(2))/((0.005)xx(0.005))=3600` or for reverse reaction `Q_(r)= 2.77xx10^(-4)`. Since `Q_(r)gt K_(SP)` of CdS, thus CdS will be precipitated. Also `Q_(r)lt K_(SP)` of ZnS, thus, ZnS will remain in solution. |
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| 1566. |
The correct order of decreasing acidic nature of `H_(2)O, ROH, CH -= CH` and `NH_(3)` isA. `CH -= CH gt H_(2)O ROH gt NH_(3)`B. `H_(2)O gt ROH gt CH -= CH gt NH_(3)`C. `ROH gt NH_(3) gt CH -= CH gt H_(2)O`D. `H_(2)O gt ROH gt NH_(3) gt CH -= CH` |
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Answer» Correct Answer - B `{:("Compound",,pK_(a)),(ROH,,(16-18)),(H_(2)O,,14),(NH_(3),,4.75 (pK_(b))),(HC -= CH,,26):}` Smaller the `pK_(a)` stronger the acid. `:.` The order of decreasing acidic nature is `H_(2)O gt ROH gt HC -= CH gt NH_(3)`. |
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| 1567. |
The solubility product of a binary weak electrolyte is `4 xx 10^(-10)` at 298 K. Its solubility in mol `dm^(-3)` at the same temperature isA. `4 xx 10^(-5)`B. `2 xx 10^(-5)`C. `8 xx 10^(-10)`D. `16 xx 10^(-20)` |
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Answer» Correct Answer - B `AB hArr A^(+) + B^(-), K_(sp) = S^(2)` `S = sqrt(K_(sp)) = sqrt(4 xx 10^(-10)) = 2 xx 1-^(-5)`. |
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| 1568. |
`pK_(a)` values of two acids A and B are 4 and 5. The strengths of these two acids are related asA. Acid A is 10 times stronger than acids BB. Strength of acid A : strength of acid B = 4:5C. The strengths of the two acids can not be comparedD. Acid B is 10 times stronger than acid A |
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Answer» Correct Answer - A `pK_(a)` of acid A =4 , `pK_(a)` of acid B = 5 We Know that `pK_(a) = -log K_(a)` `:.` Acid A `K_(a) = 10^(-4)`, Acid `B K_(a) = 10^(-5)` Hence A is ten times stronger than that of B. |
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| 1569. |
The strengths of acids and bases are directly related to their strengths as electrolytes. The electrical conductivity of `0.1 MCHl :`A. is higher than `0.1 M CH_(3)COOH`B. is lower than `0.1M CH_(3)COOH`C. equal to `0.1 M CH_(3)COOH`D. None |
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Answer» Correct Answer - 1 Higher, HCl dissociates into its ions to a much greater extent than `CH_(3)COOH`. |
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| 1570. |
`pK_(a)` values of four acids are given below at `25^(@)C` . Indicate the strongest acidA. `2.0`B. `2.5C. `3.0`D. `4.0` |
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Answer» Correct Answer - A Lesser the `pK_(a)` value , stronger the acid is . |
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| 1571. |
`pK_(a)` values of two acids. A and B are 4 and 5. The strengths of these two acids are related asA. Acid A is 10 times stronger than acid BB. Strength of acid `A : ` Strength of acid `B = 4 :5`C. The strength of two acids cannot be comparedD. Acid B is 10 times stronger than acid A. |
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Answer» Correct Answer - A Factual question. |
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| 1572. |
The Lewis acidic strength of `SO_(3)` when compared to `SO_(2)` isA. EqualB. LessC. MoreD. Can not be predicted |
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Answer» Correct Answer - C lewis acids strenght of `SO_(3)` of `SO_(2)` |
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| 1573. |
What will be the sum of pH and pOH in an aqueous solutionA. 7B. `pk_(w)`C. ZeroD. 1 |
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Answer» Correct Answer - B `pH + pOH = pK_(w)` |
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| 1574. |
pH values of HCl and NaOH solutions each of strength `(N)/(100)` will be respectivelyA. 2 and 3B. 2 and 12C. 12 and 2D. 2 and 10 |
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Answer» Correct Answer - B `(N)/(100) = 0.01 N HCl, [H^(+)] = 10^(-2) M , pH =2` `[OH] = 10^(-2) M` for NaOH `pH + pOH = 14, pH =14-2, pH = 12`. |
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| 1575. |
Find moles of `NH_(4)Cl` required to prevent `Mg(OH)_(2)` from precipitating in a litre of solution which contains 0.02 mole `NH_(3)` and 0.001 mole `Mg^(2+)` ions. Given : `K_(b)(NH_(3))=10^(-5),` `K_(sp)[Mg(OH)_(2)]=10^(-11)`.A. `10^(-4)`B. `2 xx 10^(-3)`C. `0.02`D. `0.1` |
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Answer» Correct Answer - B `[OH^(-)]= sqrt((K_(SP))/([Mg^(2+)])) = sqrt((10^(-11))/(10^(-3))) = 10^(-4)` `rArr pOH = pK_(b) + "log" ([NH_(4)Cl])/([NH_(3)])` |
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| 1576. |
The dissociation constant of water is `1 xx 10^(-14) mol ^(-2) litre ^(-2)`. What is the pH of 0.001 M KOH soluiton ?A. `10^(-11)`B. `10^(-3)`C. 3D. 11 |
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Answer» Correct Answer - D `[OH^(-)]=[KOH]=0.001M =1xx10^(-3)M` `:. p(OH)=-log (1xx 10^-3))=3` `pH=14-p(OH)=14-3=11` |
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| 1577. |
Given that the dissociation constant for `H_(2)O` is `K_(w) = 1 xx 10^(-14) "mole"^(2)//"litre"^(2)`. What is the pH of a 0.001 molar KOH solutionA. `10^(-11)`B. 3C. 14D. 11 |
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Answer» Correct Answer - D 0.001 M KOH solution `[OH^(-)] = 0.001 M = 1 xx 10^(-3) M` `[H^(+)] xx [OH^(-)] = 1 xx 10^(-14)` `[H^(+)] = (1 xx 10^(-14))/([OH^(-)])` `[H^(+)] = (1 xx 10^(-14))/(1 xx 10^(-3)) = 1 xx 10^(-14) xx 10^(+3)` `[H^(+)] = 10^(-11) M` `pH = 11`. |
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| 1578. |
A solution of `HCl` contains `0.1920 g` of an acid in `0.5 litre` of a solution. The degree of dissociation is `95%`. The `pH` of the solution isA. 2B. 1C. 3D. 4 |
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Answer» Correct Answer - A No. of moles of HCl `=(0.192)/(36.5)=0.0053` Volume `=0.5 ` lt. Molarity of HCl `=(0.0053)/(0.5)=1.06xx10^(-2)M` `[H^(+)]` in HCl solution `=1.06 xx 10^(-2) xx (95)/(100)` `~~100.7xx10^(-4)~~1xx10^(-2)` Thus `pH=- log [H^(+)]` `=- log (1xx10^(-2))=2` |
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| 1579. |
Identify the correct order of acidic strength of `CO_(2),CuO, CaO` and `H_(2)O`.A. `CaOltCuOltH_(2)OltCO_(2)`B. `H_(2)OltCuOltCaOltCO_(2)`C. `CaOltH_(2)OltCuOltCO_(2)`D. `H_(2)OltCO_(2)ltCaOltCuO` |
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Answer» Correct Answer - A CaO is strongest base, `CO_(2)` is acidic. Also CuO is strong base than `H_(2)O`. |
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| 1580. |
For the precipitation of cations of group four in the qualitative analysis, the medium is made alkaline before passing `H_(2)S` gas. The purpose of alkaline solution isA. to supress the ionisation of `H_(2)S`B. to increase the ionisation of `H_(2)S`C. to increase the ionisation of metal saltD. to decrease the ionisation of metal salt. |
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Answer» Correct Answer - B `H_(2)S hArr 2H^(+) +S^(-2)` `NH_(4)OH rarr NH_(4)^(+) +OH^(-)` `H^(+)` combines with `OH^(-)` to form water and thus the first equilibrium shifts in the forward direction to increase the ionisation of `H_(2)S`. |
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| 1581. |
Which of the following are true for an acid- base titration?A. Indicators catalyse the acid-base reactions by relasing or accepting `H^(o+)` ions.B. Indicators do not significantly affect the `pH` of the solution to which they are addedC. Acid-base reactions do not occur in the absence of indicatorsD. Indicators have different colours in dissociated and undissociated forms. |
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Answer» Correct Answer - B::D a. Wrong statement. b. Correc: Indicators are weak acid or base and are added is small amount. So they do not affect the `pH` of the solution. c. Wrong: They occur even in the absence of indicators. d. Correct: Factual statement. |
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| 1582. |
Excess of `Ag_(2)SO_(4)(s), BaSO_(4)(s)`, and `Ba_(3)(PO_(4))_(2)(s)` are simultaneously in euqilibrium with distilled water. Which of the following is (are) true? Assume no hydrolysis of dissolved ions.A. `[Ag^(o+)] +2[Ba^(2+)] = 2[SO_(4)^(2-)] +3 [PO_(4)^(3-)]`B. `2[Ag^(o+)] +4[Ba^(2+)] = 2[SO_(4)^(2-)]+2[PO_(4)^(3-)]`C. `2[Ag^(o+)]+3[Ba^(2+)]=2[SO_(4)^(2-)] + 2[PO_(4)^(3-)]`D. `[Ag^(o+)] +[Ba^(2+)] =[SO_(4)^(2-)] +[PO_(4)^(3-)]` |
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Answer» Correct Answer - A `{:(i.Ag_(2)SO_(4),hArr,2Ag^(o+)+,SO_(4)^(2-),),(,,2x,x,),(ii.BaSO_(4),hArr,Ba^(2+)+,SO_(4)^(2-),),(,,y,y,),(iii.Ba_(3)(PO_(4))_(2),hArr,3Ba^(2+)+,2PO_(4)^(3-),),(,,3z,2z,):}` `:. [Ag^(o+)] = 2x, [SO_(4)^(2-)] = x+y, [Ba^(2+)] = x+3z, [PO_(4)^(3-)] = 2z` Direct method: By electroneutrality. `[["Total positive charge"="Total negative charge",],["Total charge"="Charge on the ion"xx"Concentration",]]` iv. `1 xx [Ag^(o+)] +2[Ba^(2+)] = 2[SO_(4)^(2-)] + 2[PO_(4)^(3-)]` Alternatively: Substituting the concentration of each ion in (iv). `2x +2(x +3z) = 2(x+y) +3(2z)` Dividing by: `x+(x+3z) =(x+y)+(3)/(2)(2z)` `([Ag^(o+)])/(2) +[Ba^(2+)] = [SO_(4)^(2-)] +(3)/(2) [PO_(4)^(3-)]` or `[Ag^(o+)] +2[Ba^(2+)] = 2[SO_(4)^(2-)] +3[PO_(4)^(3-)]` Hence answer is (a) |
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| 1583. |
A solution is found to contain `[CI^(Theta)] = 1.5 xx 10^(-1)M, [Br^(Theta)] = 5.0 xx 10^(-4)M, [CrO_(4)^(2-)] = 1.9xx 10^(-2)M`. A solution of `AgNO_(3)` (100 % dissociated) is added to the above solution drop by drop. Which silver salt will precipiate first ? Given: `K_(sp) (AgCI) = 1.5 xx 10^(-10), K_(sp) (AgBr) = 5.0 xx 10^(-13),K_(sp) (Ag_(2)CrO_(4)) =1.9 xx 10^(-12)`.A. `AgCI`B. `AgBr`C. `Ag_(2)CrO_(4)`D. `AgCI` and `AgBr` togther |
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Answer» Correct Answer - D For precipitation, `[Ag^(o+)]` ion should be minimum in the solution. For `AgCI: [Ag^(o+)]_(min) = (K_(sp)AgCI)/([CI^(Theta)]) = (1.5 xx 10^(-10))/(1.5 xx 10^(-1)) = 10^(-9)M` For `AgBr: [Ag^(o+)]_(min) = (K_(sp)AgBr)/([Br^(Theta)]) = (5.0xx10^(-13))/(5.0xx10^(-4)) = 10^(-9)M` For `Ag_(2)CrO_(4): [Ag^(o+)]_(min)` `= ((K_(sp)Ag_(2)CrO_(4))/([CrO_(4)^(2)]))^((1)/(2)) = ((1.9xx10^(-12))/(1.9xx10^(-2)))=10^(-5)M` Therefore, `[Ag^(o+)]` min in solution is in `AgCI` and `AgBr`, so both will be precipitated. |
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| 1584. |
Using `CO_(2), NH_(3),NH_(4)NO_(3)`, and `K_(2)CrO_(4)` as the only reagents, devise a qualitative analysis scheme for separating and identifying the following ions, which might all be present in the same mixture: `Ba^(2+), Ca^(2+), Mg^(2+), Na^(o+), Pb^(2+)`. Assume that each cation present is `0.10M`. Sate the conditions of `pH` and the reagent concentration which are required in each step. |
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Answer» `{:(Ba^(2+)),(Ca^(2+)),(Mg^(2+)),(Na^(o+)),(Pb^(2+)):}}underset(NH_(3)+NH_(4)^(o+))(_)underset(Pb(OH)_(2)(s))({{:(Ba^(2+)),(Ca^(2+)),(Mg^(2+)),(Na^(o+)):}})underset(NH_(3))(_){{:(Ba^(2+)),(Ca^(2+)),(Na^(o+)),(Mg(OH)_(2)(s)):}}underset(NH_(3))overset(CO_(2))(_){{:(BaCO_(3)(s)),(CaCO_(3)(s)),(Na^(o+)):}` `{:(BaCP_(3),,,,),(CaCO_(3),,,,):}}underset("acidity")rarrunderset(CrO_(4)^(2-))rarr{:(BaCrO_(4)(s),,,,),(Ca^(2+)(aq),,,,):}` The `pB^(2+)` ions is identified by acidified followed by treatement with chromate ion, upon which it forms a yellow solid precitate. The `Na^(o+)` is identifed by the intense yeellow flame it yields. |
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| 1585. |
What is general criteria of chossing a suitable indicator for a given titration?A. The indicator should have a broad `pH` range.B. `pH` at the end point of titration should be close of neutral point of indicatorC. The indicator should have neutral point at `pH = 7`.D. The indicator must show a sharp colour changes near the equivalence point of titration point. |
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Answer» Correct Answer - B::D a. Wrong: Indicator should have a small transition range. b. Correct: `pH` at equivalence point `~~pH_(Ind)` c. Wrong: `pH` at equivalence point and at the end point may be different. d. Correct: Factual statement. |
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| 1586. |
`HgCrO_(4)` just begins to peripitate when equal volumes of `4 xx 10^(-4)M Hg_(2)(NO_(3))_(2)` and `2 xx 10^(-5)M K_(2)CrO_(4)` are combind. What is the approximate `K_(sp)` value of `Hg_(2)CrO_(4)`?A. `1 xx 10^(-18) mol L^(-1)`B. `8 xx 10^(-9)mol L^(-1)`C. `2xx 10^(-9) mol L^(-1)`D. `4 xx 10^(-9) mol L^(-1)` |
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Answer» Correct Answer - B `Hg_(2)CrO_(4)hArr Hg_(2)^(2+) +CrO_(4)^(2-)` `[Hg_(2)^(2+)]= 4 xx 10^(-4)M, [CrO_(4)^(2-)] = 2xx 10^(-5)` `K_(sp) = (4 xx 10^(-4)) (2xx 10^(-5)) = 8 xx 10^(-9)` |
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| 1587. |
The species present in solution when ` CO-2` is dissolved in water are :A. `CO_(2),H_(2)CO_(3),HCO_(3)^(-),CO_(3)^(2-)`B. `H_(2)CO_(3),CO_(3)^(2-)`C. `CO_(3)^(2-),HCO_(3)^(-)`D. `CO_(2),H_(2)CO_(3)` |
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Answer» Correct Answer - A All the following equilibrium exist together. `CO_(2)+H_(2)OhArrH_(2)CO_(3)` `H_(2)CO_(3)hArrH^(+)+HCO_(3)^(-)` `HCO_(3)^(-)hArr H^(+)+CO_(3)^(2-)` |
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| 1588. |
`HX` is a weak acid `(K_(a) = 10^(-5))`. If forms a salt `NaX(0.1M)` on reacting with caustic soda. The degree of hydrlysis of `NaX` isA. `0.01%`B. `0.001%`C. `0.1%`D. `0.5%` |
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Answer» Correct Answer - A For a salt of weak acid and strong base. degree of hydrolysis `= sqrt((K_(w))/(CK_(a))) = sqrt((10^(-14))/(0.1xx10^(-5))) = 10^(-4)` Precentage of hydrolysis `= 10^(-4) xx 100 = 0.01%` |
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| 1589. |
`HX` is a weak acid `(K_(a) = 10^(-5))`. If forms a salt `NaX(0.1M)` on reacting with caustic soda. The degree of hydrlysis of `NaX` isA. `0.0001%`B. `0.01%`C. `0.1%`D. `0.15%` |
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Answer» Correct Answer - B NaX is salt of weak acid with strong base. `h= sqrt((K_(H))/(C ))=sqrt((K_(w))/(K_(a).C))` `= sqrt((10^(-14))/(10^(-5)xx0.1))=sqrt(10^(-8))=10^(-4)` or `h= 10^(-4)xx100= 10^(-2)= 0.01%` |
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| 1590. |
A weak acid `HX(K_(a) = 10^(-5))` on reaction with `NaOH` gives `NaX`. For `0.1M` aqueous solution of `NaX`, the `%` hydrolysis isA. `1%`B. `0.01%`C. `0.001%`D. `0.15%` |
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Answer» Correct Answer - B Salt of `W_(A)//S_(B)` is formed `{:(X^(Theta)+,H_(2)OhArr,HX+,overset(Theta)OH,),(1,-,0,0,),(C(1-h),,Ch,Ch,):}` `:. H = sqrt((K_(h))/(c)) = sqrt((K_(w))/(K_(a).C))` `= ((10^(-14))/(10^(-5)xx0.1))^(1//2) = 10^(-4)` `% h = 10^(-4) xx 100 = 10^(-2) = 0.01` |
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| 1591. |
The hydrolysis constant `K_(h)` of a salt of `NaOH` and a weak acid (HX) if the Ka of the acid is `2 xx 10^(-6) `isA. `5 xx 10^(-8)`B. `5 xx 10^(-6)`C. `5 xx 10^(-9)`D. `2.5 xx 10^(-7)` |
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Answer» Correct Answer - C `K_(b) = (K_(w))/(Ka)` |
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| 1592. |
A weak acid `HX` has the dissociation constant `1 xx 10^(-5)M`. It forms a salt `NaX` on reaction with alkali. The percentage hydrolysis of `0.1M` solution of `NaX` isA. `0.0001%`B. `0.01%`C. `0.1 %`D. `0.15%` |
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Answer» Correct Answer - B For a salt of weak acid and strong base, degree of hydrolysis , `h=sqrt((K_(w))/(K_(a)c))=sqrt((10^(-14))/(1 xx 10^(-5)xx 0.1))` `=sqrt((10^(-14))/(10^(-6)))=10^(-4)` `%h=10^(-4)xx100=10^(-2)=0.01%` |
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| 1593. |
A weak acid `HX` has the dissociation constant `1 xx 10^(-5)M`. It forms a salt `NaX` on reaction with alkali. The percentage hydrolysis of `0.1M` solution of `NaX` isA. `0.0001%`B. `0.01%`C. `0.1%`D. `0.15%` |
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Answer» `NaX` is a salt of weak acid and strong base. :.Hydrolysis constant of `NaX` is `K_(h)=(K_(w))/(K_(a))=(1xx10^(-14))/(10^(-5))=1xx10^(-9)` For this type of salt ,`K_(h)=Ch^(2)` `C`=molar concentration `h`=degree of hydrolysis ` :.1xx10^(-9)=0.1xxh^(2)` or `h^(2)=(1xx10^(-9))/(0.1)=1xx10^(-4)` :. percentage hydrolusis of `NaX` salt =`1xx10^(-4)xx100=1xx10^(-2)=0.01%` |
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| 1594. |
Calculate the hydrolysis constant of a salt of week acid `(K_(a) = 2 xx 10^(-6))` and of a weak base `(K_(b) = 5 xx 10^(-7))`A. `10^(-4)`B. `10^(-2)`C. `10^(-6)`D. `10^(-8)` |
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Answer» Correct Answer - B `K_(h) = (kw)/(k_(a).k_(b)) = (10^(-14))/(2 xx 10^(-6) xx 5 xx 10^(-7))` |
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| 1595. |
MX is the salt of weak base, MOH and weak acid, HX. Aqueous solution of MX isA. Acidic , if `K_(a) gt K_(b)`B. Basic, if `K_(a) lt K_(b)`C. Neutral, ig `K_(a) = K_(b)`D. All the above |
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Answer» Correct Answer - D `p^(H)` is salt of `W.A` + `W.B` depends of `K_(a)` & `K_(b)` |
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| 1596. |
At `25^(@)C` the pH value of a solution is 6. The solution isA. BasicB. AcidicC. NeutralD. Both (b) and (c) |
| Answer» Correct Answer - B | |
| 1597. |
`NaOH_((aq)), HCl_((aq))` and `NaCl_((aq))` concentration of each is `10^(-3)` M. Their pH will be respectivelyA. 10, 6, 2B. 11, 3, 7C. 10, 2, 6D. 3, 4, 7 |
| Answer» Correct Answer - B | |
| 1598. |
`M_(2)SO_(4) (M^(o+)` is a monovalent metal ion) has a `K_(sp)` of `3.2 xx 10^(-6)`at `298 K`. The maximum concentration of `SO_(4)^(2-)` ion that could be attained in a saturated solution of this solid at `298 K` isA. `3 xx 10^(-3)M`B. `7 xx 10^(-2)M`C. `2.89 xx 10^(-4)M`D. `2xx10^(-2)M` |
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Answer» Correct Answer - D `{:(M_(2)SO_(4)hArr,2M^(o+)+,SO_(4)^(2-),,),(,2S,S,,):}` `K_(sp) = (2S)^(2) (S) = 4S^(3)`. `:. S =3sqrt((K_(sp))/(4))=3 sqrt((3.2xx10^(-5))/(4)) = 2xx10^(-2)M` |
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| 1599. |
`M_(2)SO_(4) (M^(o+)` is a monovalent metal ion) has a `K_(sp)` of `3.2 xx 10^(-6)`at `298 K`. The maximum concentration of `SO_(4)^(2-)` ion that could be attained in a saturated solution of this solid at `298 K` isA. `3xx10^(-3) M`B. `7xx 10^(-2) M`C. `2.89 xx 10^(-4) M`D. `2xx10^(-2)M` |
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Answer» Correct Answer - D The maximum concentration of `SO_(4)^(2-)` will be same as the solubility of `M_(2)SO_(4)`. If solubility is x mol `L^(-1)`. `M_(2)SO_(4) hArr 2M^(+)+SO_(4)^(2-)` `K_(sp)=[M^(+)]^(2)[SO_(4)^(2-)]=4x^(3)` `x=(K_(sp)//4)^(1//3)` `[SO_(4)^(2-)]=x=((K_(sp))/(4))^(1//3)` `=((3.2xx10^(-5))/(4))^(1//3)=2xx10^(-2)M` |
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| 1600. |
Assertion (A): `pH` of water increases with an increase in temperature. Reason (R) : `K_(w)` or water increases with increase in temperature.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 - D | |