1869 + Interview Questions in IONIC-EQUILIBRIUM IN GENERAL AWARENESS Page 1 InterviewSolution

1.	100ml of 0.1M `H_(3)PO_(4)` is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution `K_(1)=10^(-3)M` `K_(2)=10^(-8)M` `K_(3)=10^(-13)M` pH at 2nd equivalent point will be :A. `5.5`B. `10.5`C. `8`D. `7`
Answer» Correct Answer - b

Discussion

2.	The conjugate bas of hydrazoic acid isA. `N_(2)H_(4)`B. `N_(2)H_(5^(+))`C. `N_(3)^(-)`D. `NH_(2)OH`
Answer» Correct Answer - C `underset("Acid")(N_(3)H) overset(-H^(+))rarr underset(C.B)(N_(3)^(-))`

Discussion

3.	Calcualte the percentage hydrolysis of `10^(-3)M N_(2)^(o+)H_(5)C1^(Theta)` (hydrazinium chloride), salt contining acid ion conjugate to hydrazine base `(NH_(2)NH_(2)). K_(b)` for `N_(2)H_(4) = 1.0 xx 10^(-6)`.
Answer» Correct Answer - B::C `[N_(2)^(o+)H_(5)]C1^(Theta)` is a salt of `W_(B)//S_(A)`. `K_(h) = (K_(w))/(K_(b)) = (10^(-14))/(10^(-6)) = 10^(-8)` `K_(h) = sqrt((K_(h))/(c)) = ((10^(-8))/(10^(-3)))^(1//2) = (10^(-5))^((1)/(2))` `= (10 xx 10^(-6))^(1//2)` `= sqrt(10) xx 10^(-3)` `= 3.162 xx 10^(-3)` `%h = 3.16 xx 10^(-3) xx 100` `= 0.316 ~~ 0.32%`

Discussion

4.	Calculate the hydrolysis constant `(K_(h))` and degree of hydrloysis `(h)` of `NH_(4)C1` in `0.1M` solution. `K_(b) = 2.0 xx 10^(-5)`. Calculate the `[overset(Theta)OH]` ions in the solution.
Answer» Correct Answer - A i. `NH_(4)C1` is a salt of `W_(B)//S_(A)`. `K_(h) = (K_(w))/(K_(b)) = (10^(-14))/(2xx10^(-5)) = 5 xx 10^(-10)`. ii. `h = sqrt((K_(h))/(c))` `= ((5 xx 10^(-10))/(0.1))^((1)/(2))` `= sqrt(50) xx 10^(-5)` `= 7.07 xx 10^(-5)` iii. `[overset(Theta)OH] = Ch = 0.1 xx 7.07 xx 10^(-5) = 7.07 xx 10^(-6)M`.

Discussion

5.	A trpical aspirin tablet constains 324 mg of aspirin (acetyl salicylic acid `C_(9)H_(8)O_(4)`) a monoprotic acid having `K_(a)=3.0xx10^(-4)` . What is the degree of dissociation of salt and pH of the solution, if two aspirin tables are dissolved to prepare 300 mL solution in water ?
Answer» Moles of aspirin= `(2xx324xx10^(-3))/(180)=3.6xx10^(-3)` `:. ["aspirin"]=(3.6xx10^(-3))/(300)xx10^(3)=0.012` `C_(9)H_(8)O_(4)hArrC_(9)H_(7)O_(4)+H^(+)` `K_(a)=(Calpha^(2))/((1-alpha))=(0.012xxalpha^(2))/((1-alpha))=3.610^(-3)` `:. 0.012alpha^(2)+0.0036alpha-0.0036=0` `alpha=(-0.0036+-sqrt((0.0036)^(2)+4xx0.0036xx0.012))/(2xx0.012)` `=(-0.0036+-sqrt(1.296xx10^(-5)+17.28xx10^(-5)))/(2xx0.012)` `=0.42` `:. [H^(+)]=Calpha=0.42xx0.012=5.04xx10^(-3)` `:. pH= -log[H^(+)]= -log 5.04xx10^(-3)` `=2.2976`

Discussion

6.	What is the entropy that takes place in conversion of 1 mol of `alpha-`tin to 1 mol `beta -` tin at `17^(@)C`, 1 atm pressure if enthalpy of transition is 2.90 kJ `mol^(-1)` ?A. `1.0 kJ mol^(-1)`B. `7.32 JK^(-1) mol^(-1)`C. `10JK^(-1)mol^(-1)`D. `100JK mol^(-1)`
Answer» Correct Answer - C `DeltaS=(DeltaH_("transition"))/(T_("transition"))=5 xx 10^(-11)mol^(-1)` `=(2.90 xx 1000J mol^(-1))/(290)=10JK^(-1)mol^(-1)`

Discussion

7.	We represent various reagents pictorially such that, The relative area of the squars represents the relative volume of the reagents used and the overlapping region represents the relative amount of mixing of the reagents. (Use log 5=0.7,log2=0.3,log3=0.5,`d_(H_2O)=1gml^(-1)`) For a solution represented by Which is/are correct?A. Ph=1B. `alpha` of weak acid =`5xx10^(-4)`C. `alpha` of `H_(2)O=9xx10^(-15)`D. Ph=1.7
Answer» Correct Answer - b,c,d

Discussion

8.	Determine the concentration of `NH_(3)` solution whose one litre can dissolve `0.10` mole AgCI. `K_(SP)` of AgCI and `K_(f)` of `Ag(NH_(3))_(2)^(+)` are `1.0xx10^(-10)M^(2)` and `1.6xx10^(7)M^(-2)` respectively.
Answer» `AgCI+2NH_(3)hArr Ag(NH_(3))_(2)^(+)+CI^(-)` `AgCI_((s))hArrAg^(+)+CI^(-)` `K_(SP)=[Ag^(+)][CI^(-)]` ………..(1) Given `Ag^(+)+2NH_(3)hArr Ag(NH_(3))_(2)^(+)` `K_(f)=([Ag(NH_(3))_(2)^(+)])/([Ag^(+)][NH_(3)]^(2))` ………….(2) By eqs. (1) and (2) `K_(SP)xxK_(f)=([Ag(NH_(3))_(2)^(+)][CI^(-)])/([NH_(3)]^(2))` or `1xx10^(-10)xx1.6xx10^(7)=(axxa)/([NH_(3)]^(2))` Given solubility of `AgCI= 0.1M` `:. a=0.1M` for `Ag(NH_(3))_(2)^(+)` and `CI^(-)` `:. [NH_(3)]^(2)=(0.1xx0.1)/(1.6xx10^(-3))=6.25` `[NH_(3)]=2.5M` Also `0.2M NH_(3)` is needed to dissolve `0.1M Ag^(+)` ionsthus `[NH_(3)]=2.5+0.2=2.7M`

Discussion

9.	If `500 ml` of `0.4 M AgNO_(3)` is mixed with `50 ml` of `2 M NH_(3)` solution then what is the concentration of `[Ag(NH_(3))]^(+)` in solution `(K_(t), [Ag_(NH_(3))]^(+) = 10^(3), K_(f_(2)) [Ag(NH_(3))_(2)]^(+)= 10^(4))`A. `3.33 xx 10^(-7) M`B. `3.33 xx 10^(-5) M`C. `3 xx 10^(-4) M`D. `10^(-7) M`
Answer» Correct Answer - B After mixing `[Ag^(+)] = 0.2M, [NH_(3)] = 1M` Due to very high value of `K_(f), Ag^(+)` is mainly converted into complex. `Ag_((aq))^(+)+2NH_(3(aq))hArr[Ag(NH_(3))_(2)]_((aq))^(+)` `{:(0.2,1,),(x,0.6,02):}` `{:([Ag(NH_(3))_(2)]_((aq))^(+)hArr,[Ag(NH_(3))]_((aq))^(+),+NH_(3(aq))),(0.2-y,y,0.6+y),(0.2,,0.6):}` `(1)/(K_(f_(2))) = (y xx 0.6)/(0.2) = (1)/(10^(4))` `y = [Ag(NH_(3))]^(+) = 3.33 xx 10^(-5) M`

Discussion

10.	a. Calculate `[Ag^(o+)]` in a solution of `[Ag^(o+)]` in a solution of `[Ag(NH_(3))_(2)^(o+)]` prepared by adding `1.0 xx 10^(-3)mol AgNO_(3)` to `1.0L of 1.0M NH_(3)` solution `K_(f) Ag(NH_(3))_(2)^(o+) = 10^(8)`. b. Calculate `[Ag^(o+)]` which is in equilibrium with `0.15M [Ag(NH_(3))_(2)]^(o+)` and `1.5 NH_(3)`.
Answer» Correct Answer - A::B i.Since `K_(f)` is so large, most of `Ag^(o+)` will form complex ions. `{:("Initial mol"),("Reacted"),("At equilibrium")]{:(Ag^(o+)+,2NH_(3)hArr,Ag(NH_(3))_(2)^(o+),,,),(10^(-3),0.1,0,,),(-x,-2x,0,,),(10^(-3)-x~~10^(-3),0.1-2x,10^(-3)-x~~10^(-3),,),(,0.1-2(10^(-3)),,,),(,~~0.098,,,):}` `K_(f) =([Ag(NH_(3))_(2)^(o+)])/([Ag^(o+)][NH_(3)]^(2))` `10^(8) = (10^(-3))/((x)(0.098)^(2)) :. x = [Ag^(o+)] = 1 xx 10^(-9)M` ii. `10^(8) = (0.15)/((x)(1.5)^(2)) :. x = [Ag^(o+)] = 7 xx 10^(-10)M`.

Discussion

11.	1 mole of `AgNO_(3)` is added to 10 litre of 1 M `NH_(3)`. What is the concentration of `Ag(NH_(3))^(+)` in solution? [Given : For `Ag(NH_(3))_(2)^(+),K_(f_(1))=2.0xx10^(3),K_(f_(2))=10^(3)`]A. `8xx10^(-5)`B. `1.25xx10^(-5)`C. `4xx10^(-5)`D. `1.25xx10^(-5)`
Answer» Correct Answer - b

Discussion

12.	Solubility of a solute in water is dependent on temperature as given by `S=Ae^(-DeltaH//RT)`, where `DeltaH`=heat of solution `"Solute"+H_(2)O(l) hArr "Solution", DeltaH= +- x` For given solution, variation of log S with temperature is shown graphically. Hence, solution is A. `CaO`B. `MgSO_(4)`C. `CuSO_(4)`D. `CuSO_(4).5H_(2)O`
Answer» Correct Answer - D In `S vs(1)/(T)` graph with negative slope. ` :. DeltaH gt 0 rArr` Salts with water of crystallisation. i.e., `CuSO_(4).5H_(2)O`

Discussion

13.	Calculate pH a) `NaH_(2)PO_(4)` b) `Na_(2)HPO_(4)` respectively, for `H_(3)PO_(4) pKa_(1) = 2.25, pKa_(2) = 7.20, pKa_(3) = 12.37`)A. `9.78,4.68`B. `4.68,4.68`C. `9.78,9.78`D. `4.68,9.78`
Answer» Correct Answer - D `pH` of `H_(2)PO_(4)^(-) =- (pKa_(1) + pKa_(2))/(2)` `pH` of `HPO_(4)^(-2) = (pKa_(2) + pKa_(3))/(2)`

Discussion

14.	pH of `0.1M Na_(2)HPO_(4)` and `0.2M NaH_(2)PO_(4)` are respectively: `(pK_(a)"for" H_(3)PO_(4)` are `2.12, 7.21` and `12.0` for respective dissociation to `HPO_(4)^(2-), HPO_(4)^(-)` and `PO_(4)^(3-))`:A. `4.7,9.6`B. `9.6,4.7`C. `9.3,4.4`D. `4.4,9.3`
Answer» For `Na_(2)HPO_(4),pH=(pK_(a_(2))+pK_(a_(3)))/(2)=(7.28+12)/(2)=9.6` For `Na_(2)HPO_(4),pH=(pK_(a_(1))+pK_(a_(2)))/(2)=(2.2+7.2)/(2)=4.7`

Discussion

15.	The addition of `NaH_(2)PO_(4)` to `0.1M H_(3)PO_(4)` will cuaseA. No change in `pH` valueB. Increases in its `pH` valueC. Decrease in its `pH` valueD. Change in `pH` but cannot be predicted
Answer» Correct Answer - B On adding salt of `W_(A)//S_(B) (NaH_(2)PO_(4))` to `0.1M H_(3)PO_(4)`, an acidic buffer is formed so `pH` increases.

Discussion

16.	What is the pH value of `1 M H_(2)SO_(4)`
Answer» Correct Answer - D `1M H_(2)SO_(4) = 2N H_(2)SO_(4)` `[H^(+)] = N. alpha = 2 xx 1 = 2` `pH = -log[H^(+)] = -log 2 = -0.3010`.

Discussion

17.	40 ml of `0.1` M ammonia is mixed with 20 ml of `0.1 M HCI`. What is the pH of the mixture ? (`pK_(b)` of ammonia solution is `4.74.`)A. 4.74B. 2.26C. 9.26D. `5.00`
Answer» Correct Answer - C 40 Ml of 0.1 M HCl `=20 xx 0.1` milli mole `=2` milli mole `NH_(4)OH+HCl rarr NH_(4)Cl +H_(2)O` 2 milli mole of HCl will neutralize 2 milli moles of `NH_(4)OH` to form 2 milli moles of `NH_(4)Cl`. `NH_(4)OH ` left `=2 ` milli moles Total volume `=60 mL` `:. [NH_(4)OH]=2//60M,[NH_(4)Cl]=2//60M` ` pOH =pK_(b)+log.([NH_(4)Cl])/([NH_(4)OH])` `=4.74 +log. (2//60)/(2//60)=4.74` `:. pH =14-4.74 =9.26`

Discussion

18.	Statement: Acidic nature of boron trihalides is in the order `BF_(3)ltBCI_(3)ltBBr_(3)ltBI_(3)`. Explanation: Basic nature of nitrogen trihalides is in the order `NF_(3)gtNCI_(3)gtNBr_(3)gtNI_(3)`.A. S is correct but E is wrong.B. S is wrong but E is correc.C. Both S and E are correct and E is correct explanation of S.D. Both S and E are correct but E is correct explanation of S.
Answer» Correct Answer - A Statement is correct due to back-bonding in boron. In nitrogen halides, the order is `NF_(3)ltNCI_(3)ltNBr_(3)ltNI_(3)` on account of decreasing electronegativity of halogents. In `NF_(3)`, the lone pair is not released easily due to more `+ve` charge on N.

Discussion

19.	A solution contains `60 mL` of `0.1 M NH_(4)OH` and `30 mL` of `0.1 MHCl`. The `P^(H)` of the resulting mixture is (Given `K_(b)` of `NH_(4)OH = 1.8 xx 10^(-5)`, `1.8 = 0.2553`)A. `4.7447`B. `3.7447`C. `9.2553`D. `12.523`
Answer» Correct Answer - C `NH_(4)OH + HCl rarr NH_(4)Cl + H_(2)O ` `[NH_(4)OH] = 60 xx 0.1 - 30 xx 0.1` `[NH_(4)Cl] = 30 xx 0.1` `:. P^(H) = 14 - P^(kb)`

Discussion

20.	40 ml of 0.1 M ammonia solution is mixed with 20 ml of 0.1 MHCl. What is the pH of the mixture (p`K_(b)` of ammonia solution is 4.74)A. 4.74B. 2.26C. 9.26D. 5
Answer» Correct Answer - C At half stage of titration [salt] = [base], thus as per Handerson equation `pOH = pK_(b)`.

Discussion

21.	In the reaction `NH_(3)+BF_(3) hArr NH_(3) rarr BF_(3), BF_(3)` isA. Lewis acidB. Lewis baseC. Neither Lewis acid not Lewis baseD. Lewis acid and Lewis base both
Answer» Correct Answer - A `BF_(3)` accept between pair form `NH_(3)` so it Lewis acid.

Discussion

22.	Assertion (A): Both reactions are Lewis acid-base recations? i. `NH_(3) + BF_(3) rarr H_(3)N: BF_(3)` ii. `Mg +S rarr Mg^(2+) +S^(2-)` Reason (R) : Lewis acid-base reaction involve the donation of lone pair electrons from base to acid. this donation results in a corrdinate bond.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.
Answer» Correct Answer - D `(A)` only reaction (i) is Lewis acid-base reaction. Hence `(A)` is wrong. `(R)` is correct.

Discussion

23.	The concentration of hydroxyl ion in a solution left after mixing 100 mL of 0.1 M `MgCl_(2)` and 100 mL of 0.2 M NaOH: (`K_(sp)` of `Mg(OH)_(2)=1.2xx10^(-11)`)A. `2.88xx10^(-3)`B. `2.88xx10^(-2)`C. `2.88xx10^(-4)`D. `2.88xx10^(-5)`
Answer» Correct Answer - c

Discussion

24.	The concentration of hydroxyl ion in a solution left after mixing 100 mL of `0.1M MgCl_(2)` and 100 mL of `0.2M NaOH` is: `(K_(SP)of Mg(OH)_(2)=1.2xx10^(-11))`A. `2.8xx10^(-3)`B. `2.8xx10^(-2)`C. `2.8xx10^(-4)`D. `2.8xx10^(-5)`
Answer» Correct Answer - C `{:(,MgCI_(2)+,2NaOHrarr,Mg(OH)_(2)+,2NaCI),("mm before", 10,20,0,0),("reaction",0,0,10,20):}` Thus, 10 milli-mole of `Mg(OH)_(2)` are formed The product of `[Mg^(2+)][OH^(-)]^(2)` is therefore `[(10)/(200)]xx[(20)/(200)]^(2)=5xx10^(-4)` which is more than `K_(SP)of Mg(OH)_(2)`. Now solubiltiy (S) of `Mg(OH)_(2)` can be derived by `K_(SP)=4S^(3)` `:. S= 3sqrt((K_(SP))/(4))=3sqrt([(1.2xx10^(-11))/(4)])` `=1.4xx10^(-4)` `:. [OH^(-)]= 2S =2.8xx10^(-4)`

Discussion

25.	How many of the following salts: i. `NH_(4)C_(2)H_(3)O_(2)` ii. `PhCOONH_(4)` iii. `NaC_(2)H_(3)O_(2)` iv. `NH_(4)CI` v. `MgS` vi. `Na_(2)SO_(4)` vii. `KCI` a. Hydrolyse more in water at `25^(@)C`. b. Do not hydrolyse. c. Both cation and anion hydrolyse to the same extent. d. Both cation and anion hydrolyse to differnet extent.
Answer» Correct Answer - A::B::C a. Salt of `W_(A)//W_(B)` and `W_(B)//S_(A)` hydrolyses more in water. III. `NaC_(2)H_(3)O_(2)` (salt of `W_(A)//S_(B)) rarr C_(2)H_(3)O_(2)^(Theta)`. IV. `NH_(4)CI` (salt of `W_(B)//S_(B)) rarr overset(o+)NH_(4)` V. `MgS` (salt of `W_(A)//S_(B)) rarr S^(2-)` b. Salt of `S_(A)//S_(B)` do not hydrolyse. VI. `Na_(2)SO_(4)` VIII. `KCI` c. Salt of `W_(A)//S_(B) (NH_(4)C_(2)H_(3)O_(2))` with same value of `K_(a)` and `K_(b)`, both hydrolyse to the same extent. `(K_(a)` of `CH_(3)COOH = K_(b) of NH_(3))` d. II. Salt of `W_(A)//W_(B)` with different `K_(a)` and `K_(b)`, both hydrolyse to the same extent. `(PhCOONH_(4)` have different `K_(a)` and `K_(b))`

Discussion

26.	At `18^(@)C` aniline and acetic acid have dissociation constants `5xx10^(-10)` and `1.8xx10^(-5)` respectively. An aqueous solution of anilium acetate is hydrolysed to the extent of `x%` under equilibrium, what is pH of the solution ? `(K_(w)=10^(-14))`
Answer» Correct Answer - `4.7219;`

Discussion

27.	Solubility of AgCl in pure water is `10^(-5)` mol/litre at `25^(@)`C then, calculate solubility of AgCl in 0.1 M aqueous solution of KCI at `25^(@)` CA. `10^(-9)` mol/litreB. `10^(-5)` mol/litreC. `10^(-7)` mol/litreD. `10^(-4)` mol/litre
Answer» Correct Answer - a

Discussion

28.	The first and second dissociation constants of an acid `H_(2)A` are `1 xx 10^(-5)` and `5 xx 10^(-10)` respectively. The overall dissociation constant of the acid will beA. `9.6 xx 10^(-6)`B. `1.4 xx 10^(-1)`C. `1.2 xx 10^(-6)`D. `1.3 xx 10^(-8)`
Answer» Correct Answer - B `K = K_(1) xx K_(2)` ltbr?gt `k = 1 xx 10^(-5) xx 5 xx 10^(-10) = 5 xx 10^(-15)`

Discussion

29.	The first and second dissociation constant of an acid `H_(2)A` are `1.0xx10^(-5)` and `5.0xx10^(-10)` repectively. The overall dissociation constant of the acid will beA. `5.0xx10^(-15)`B. `0.2xx10^(5)`C. `5.0xx10^(-5)`D. `5.0xx10^(15)`
Answer» Correct Answer - a

Discussion

30.	The first and second dissociation constant of an acid `H_(2)A` are `1.0xx10^(-5)` and `5.0xx10^(-10)` repectively. The overall dissociation constant of the acid will beA. `5.0xx10^(-5)`B. `5.0xx10^(15)`C. `5.0xx10^(-15)`D. `0.2xx10^(5)`
Answer» Correct Answer - C `H_(2)AhArrHA^(-)+H^(+)` `K_(1) = ([HA^(-)][H^(+)])/([H_(2)A])` ……(1) `HA^(-)hArr H^(+)+A^(2-)` `K_(2) = ([H^(+)][A^(2-)])/([HA^(-)])` ……(2) For the reaction, `H_(2)A hArr 2H^(+)+A^(2-)` `K = ([HA^(-)][H^(+)])/([H_(2)A])xx([H^(+)][A^(2-)])/([HA^(-)])` `= ([H^(+)]^(2-)[A^(2-)])/([H_(2)A])= K_(1)xxK_(2)` `1xx10^(-5)xx5xx10^(-10)` `=5xx10^(-15)`

Discussion

31.	The solubility product of AgCl under standard conditions of temperature is given byA. `1.6 xx 10^(-5)`B. `1.5 xx 10^(-8)`C. `3.2 xx 10^(-10)`D. `1.5 xx 10^(-10)`
Answer» Correct Answer - D

Discussion

32.	The first and second dissociation constants of an acid `H_(2)A` are `1.0 xx 10^(-5)` and `5.0 xx 10^(-10)` respectively. The overall dissociation constant of the acid will beA. `5.0 xx 10^(-5)`B. `5.0 xx 10^(15)`C. `5.0 xx 10^(-15)`D. `0.0 xx 10^(-5)`
Answer» Correct Answer - C `H_(2)A hArr HA^(-) + H^(+) " " K_(1) = ([HA^(-)][H^(+)])/([H_(2)A])` ...(i) `HA^(-) rarr H^(-) + A^(2-) " " K_(2) = ([H^(+)][A^(2-)])/([HA^(-)])` ...(ii) For the reaction `H_(2)A hArr 2H^(+) + A^(2-)` `K = ([H^(+)]^(2) [A^(2-)])/([H_(2)A]) = K_(1) xx K_(2)` `= 1 xx 10^(-5) xx 5 xx 10^(-10) = 5 xx 10^(-15)`.

Discussion

33.	Solubility product of silver bromide is `5.0 xx 10^(-13)`. The quantity of potassium bromide (molar mass taken as `120 g mol^(-1)`) to be added to 1 litre of 0.5 M solution of silver nitrate to start the precipitation of AgBr isA. `5.0 xx 10^(-8) g`B. `1.2 xx 10^(-10) g`C. `1.2 xx 10^(-9) g`D. `6.2 xx 10^(-5) g`
Answer» Correct Answer - C `Br^(-) = (K_(sp)(AgBr))/(C_(Ag^(+)))` `Br^(-) = (5 xx 10^(-13))/(0.05) = 10^(-11)` Conc. `= [KBr] = 10^(-11)` ltbgt Moles of KBr `= 10^(-11)` Weight of KBr `= 10^(-11) xx 120 xx 1.2 xx 10^(-9)g`.

Discussion

34.	In which of the following solution, solubility of AgCl will be maximum?A. 0.1 M `AgNO_(3)`B. WaterC. 0.1 M `-NH_(3)(aq)`D. 0.1 M NaCl
Answer» Correct Answer - c

Discussion

35.	Solid `Ba(NO_(3))_(2)` is gradually dissolved in a `1.0 xx 10^(-4) M Na_(2)CO_(3)` solution. At what concentration of `Ba^(2+)` will a precipitate begin to form (`K_(sp)` for `BaCO_(3) = 5.1 xx 10^(-9)`)A. `4.1 xx 10^(-5) M`B. `5.1 xx 10^(-5) M`C. `8.1 xx 10^(-8) M`D. `8.1 xx 10^(-7) M`
Answer» Correct Answer - B `{:(Na_(2)CO_(3),rarr,2Na^(+),+,CO_(3)^(-2)),(1xx 10^(-4)M,,2 xx 10^(-4)M,,1 xx 10^(-4)M):}` `K_(sp)[BaCO_(3)] = [Ba^(+2)][CO_(3)^(-2)]` `5.1 xx10^(-9) = [Ba^(+2)] xx 1 xx 10^(-4)` `[Ba^(+2)] = 5.1 xx 10^(-5) M`.

Discussion

36.	In an aqueous solution `10^(-2)` M `Na_(2)SO_(4)` and `10^(-2)` M NaI are present. Now pure `Pb(NO_(3))_(2)` is added gradually then calculate concentration of `SO_(4)^(2-)` when `PbI_(2)` start precipitating in solution [`K_(sp) (PbI_(2))=10^(-9)` and `K_(sp)(PbSO_(4))=10^(-8)`].A. `10^(-2)` MB. `10^(-3)` MC. `10^(-6)` MD. `10^(-5)` M
Answer» Correct Answer - b

Discussion

37.	What is the `[HCOO^-]` in the solution that contains 0.015 M HCOOH and 0.02 M HCl ? `K_a(HCOOH)=1.8xx10^(-4)`A. `1.8xx10^(-4)`B. `1.35xx10^(-4)`C. `1.8xx10^(-2)`D. `8xx10^(-3)`
Answer» Correct Answer - b

Discussion

38.	A 0.015 M solution of a weak acid has a pH of 3.52 . What is the value of the `K_a` for this acid?A. `2.0xx10^(-2)`B. `6.2xx10^(-6)`C. `9.1xx10^(-8)`D. `1.4xx10^(-9)`
Answer» Correct Answer - b

Discussion

39.	Calculate the pH of a 0.10 N solution of `H_2CO_3` A. 3.68B. 5.76C. 7.36D. 9.34
Answer» Correct Answer - a

Discussion

40.	A 0.10 M solution of weak acid is 5.75% ionized. What is the `K_a` value for this acid ?A. `3.3xx10^(-3)`B. `3.3xx10^(-4)`C. `4.2xx10^(-5)`D. `3.3xx10^(-5)`
Answer» Correct Answer - b

Discussion

41.	Calculate the change in pH when a 0.1 M solution of `CH_3COOH` in water at `25^@C` is diluted to a final concentration of 0.01 M . `[K_a=1.85xx10^(-5)]`A. `+0.5`B. `+0.4`C. `+0.47`D. `+0.6`
Answer» Correct Answer - a

Discussion

42.	What is the aq. Ammonia concentration of a solution prepared by dissolving 0.15 mole of `NH_4^+CH_3COO^-` in 1 L `H_2O` ? `[K_(a(CH_3COOH))=1.8xx10^(-5)=K_(b(NH_3))]`A. `8.3xx10^(-4)`B. 0.15C. `6.4xx10^(-4)`D. `3.8xx10^(-4)`
Answer» Correct Answer - a

Discussion

43.	A 0.052 M solution of benzoic acid, `C_(6)H_(5)COOH`, is titrated with a strong base. What is the `[H^(+)]` of the solution one-half way to the equivalence point ? `{:("Equilibrium","Constant,"K_(a)),(C_(6)H_(5)COOH,6.3xx10^(-5)):}`A. `6.3xx10^(-5) M`B. `1.8xx10^(-3) M`C. `7.9xx10^(-3) M`D. `2.6xx10^(-2) M`
Answer» Correct Answer - a

Discussion

44.	The solubility product of AgCl is `1.8xx10^(-10)`. Precipitation of AgCl will occur only when equal volumes of solutions of :A. `10^(-8)MAg^(+)and10^(-8)MCl^(-)`ionsB. `10^(-3)MAg^(+)and10^(-3)MCl^(-)`ionsC. `10^(-6)MAg^(+)and10^(6)MCl^(-)`ionsD. `10^(-10)MAg^(+)and10^(-10)MCl^(-)`ions
Answer» Correct Answer - 2

Discussion

45.	Both conc. Being 1 M, the factor by which the percent hydrolysis of `CH_3COO^-` is greater in `CH_3COONH_4` than in `CH_3COONa` is : `[K_a(CH_3COOH)=K_b(NH_4OH)=1.8xx10^(-5)]`A. 236B. 136C. 36D. 3
Answer» Correct Answer - a

Discussion

46.	The solubility product of AgCl is `1.8xx10^(-10)`. Precipitation of AgCl will occur only when equal volumes of solutions of :A. `10^(-4) M Ag^(+) " and " 10^(-4) M Cl^(-)` are mixedB. `10^(-7) M Ag^(+) " and " 10^(-7) M Cl^(-)` are mixedC. `10^(-5) M Ag^(+) " and " 10^(-5) M Cl^(-)` are mixedD. `2xx10^(-5) M Ag^(+) " and " 2xx10^(-5) M Cl^(-)` are mixed
Answer» Correct Answer - a

Discussion

47.	`2g` of `NaOH` per `250 mL` of solution is added to a buffer solution of buffer capacity `0.2`. Then the change in pH isA. `0.5`B. `1`C. `1.5`D. `2.0`
Answer» Correct Answer - B Buffer capcity `= ("No. of moles of an acid added per lit solution")/(P^(H) "change")` `P^(H)` change `= ((8)/(40))/(0.2) = (0.2)/(0.2) = 1`

Discussion

48.	Buffer capacity of a buffer solution is `x`, the volume of `1 M NaOH` added to `100 mL` of this solution if the change of `pH` by `1` isA. 0.1 x mLB. 10 x mLC. 100 x mLD. x mL
Answer» Correct Answer - c

Discussion

49.	The volume of water that must be added to `100 ml` of `NaOH` solution to change its pH value from `12.6990` to `12` isA. `500 mL`B. `400 mL`C. `100 mL`D. `200 mL`
Answer» Correct Answer - B (i) `pH = 12.699, [OH^(-)] = 5 xx 10^(-2)` `pOH = 1.3010` (ii) `pH = 12, pOH = 2, [OH^(-)] = 10^(-2)` `100 xx 5 xx 10^(-2) = 10^(-2) xx V` `V = 500` Volume of water acid `= 500 - 100` `= 400 ml`

Discussion

50.	The solubility product of `BaCrO_(4)` is `2.4xx10^(-10)M^(2)`. The maximum concentration of `Ba(N0_(3))_(2)` possible without precipitation in a `6xx10^(-4)` M `K_(2)CrO_(4)` solution is :A. `4xx10^(-7) M`B. `1.2 xx 10^(-10)` MC. `6xx10^(-4)` MD. `3xx10^(-4)M`.
Answer» Correct Answer - A `[Ba^(2+)][CrO_(4)^(2-)]=K_(sp)` `[Ba^(2+)]=K_(sp)//[CrO_(4)^(2-)]` `[CrO_(4)^(2-)]=[K_(2)CrO_(4)]=6xx10^(-4)M` `=(2.4 xx 10^(-10))/(6 xx 10^(-4))=4 xx 10^(-7)M`

Discussion

Explore topic-wise InterviewSolutions in .

100ml of 0.1M `H_(3)PO_(4)` is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution `K_(1)=10^(-3)M` `K_(2)=10^(-8)M` `K_(3)=10^(-13)M` pH at 2nd equivalent point will be :A. `5.5`B. `10.5`C. `8`D. `7`

The conjugate bas of hydrazoic acid isA. `N_(2)H_(4)`B. `N_(2)H_(5^(+))`C. `N_(3)^(-)`D. `NH_(2)OH`

Calcualte the percentage hydrolysis of `10^(-3)M N_(2)^(o+)H_(5)C1^(Theta)` (hydrazinium chloride), salt contining acid ion conjugate to hydrazine base `(NH_(2)NH_(2)). K_(b)` for `N_(2)H_(4) = 1.0 xx 10^(-6)`.

Calculate the hydrolysis constant `(K_(h))` and degree of hydrloysis `(h)` of `NH_(4)C1` in `0.1M` solution. `K_(b) = 2.0 xx 10^(-5)`. Calculate the `[overset(Theta)OH]` ions in the solution.

A trpical aspirin tablet constains 324 mg of aspirin (acetyl salicylic acid `C_(9)H_(8)O_(4)`) a monoprotic acid having `K_(a)=3.0xx10^(-4)` . What is the degree of dissociation of salt and pH of the solution, if two aspirin tables are dissolved to prepare 300 mL solution in water ?

What is the entropy that takes place in conversion of 1 mol of `alpha-`tin to 1 mol `beta -` tin at `17^(@)C`, 1 atm pressure if enthalpy of transition is 2.90 kJ `mol^(-1)` ?A. `1.0 kJ mol^(-1)`B. `7.32 JK^(-1) mol^(-1)`C. `10JK^(-1)mol^(-1)`D. `100JK mol^(-1)`

Determine the concentration of `NH_(3)` solution whose one litre can dissolve `0.10` mole AgCI. `K_(SP)` of AgCI and `K_(f)` of `Ag(NH_(3))_(2)^(+)` are `1.0xx10^(-10)M^(2)` and `1.6xx10^(7)M^(-2)` respectively.

If `500 ml` of `0.4 M AgNO_(3)` is mixed with `50 ml` of `2 M NH_(3)` solution then what is the concentration of `[Ag(NH_(3))]^(+)` in solution `(K_(t), [Ag_(NH_(3))]^(+) = 10^(3), K_(f_(2)) [Ag(NH_(3))_(2)]^(+)= 10^(4))`A. `3.33 xx 10^(-7) M`B. `3.33 xx 10^(-5) M`C. `3 xx 10^(-4) M`D. `10^(-7) M`

1 mole of `AgNO_(3)` is added to 10 litre of 1 M `NH_(3)`. What is the concentration of `Ag(NH_(3))^(+)` in solution? [Given : For `Ag(NH_(3))_(2)^(+),K_(f_(1))=2.0xx10^(3),K_(f_(2))=10^(3)`]A. `8xx10^(-5)`B. `1.25xx10^(-5)`C. `4xx10^(-5)`D. `1.25xx10^(-5)`

Calculate pH a) `NaH_(2)PO_(4)` b) `Na_(2)HPO_(4)` respectively, for `H_(3)PO_(4) pKa_(1) = 2.25, pKa_(2) = 7.20, pKa_(3) = 12.37`)A. `9.78,4.68`B. `4.68,4.68`C. `9.78,9.78`D. `4.68,9.78`

pH of `0.1M Na_(2)HPO_(4)` and `0.2M NaH_(2)PO_(4)` are respectively: `(pK_(a)"for" H_(3)PO_(4)` are `2.12, 7.21` and `12.0` for respective dissociation to `HPO_(4)^(2-), HPO_(4)^(-)` and `PO_(4)^(3-))`:A. `4.7,9.6`B. `9.6,4.7`C. `9.3,4.4`D. `4.4,9.3`

The addition of `NaH_(2)PO_(4)` to `0.1M H_(3)PO_(4)` will cuaseA. No change in `pH` valueB. Increases in its `pH` valueC. Decrease in its `pH` valueD. Change in `pH` but cannot be predicted

What is the pH value of `1 M H_(2)SO_(4)`

40 ml of `0.1` M ammonia is mixed with 20 ml of `0.1 M HCI`. What is the pH of the mixture ? (`pK_(b)` of ammonia solution is `4.74.`)A. 4.74B. 2.26C. 9.26D. `5.00`

A solution contains `60 mL` of `0.1 M NH_(4)OH` and `30 mL` of `0.1 MHCl`. The `P^(H)` of the resulting mixture is (Given `K_(b)` of `NH_(4)OH = 1.8 xx 10^(-5)`, `1.8 = 0.2553`)A. `4.7447`B. `3.7447`C. `9.2553`D. `12.523`

40 ml of 0.1 M ammonia solution is mixed with 20 ml of 0.1 MHCl. What is the pH of the mixture (p`K_(b)` of ammonia solution is 4.74)A. 4.74B. 2.26C. 9.26D. 5

In the reaction `NH_(3)+BF_(3) hArr NH_(3) rarr BF_(3), BF_(3)` isA. Lewis acidB. Lewis baseC. Neither Lewis acid not Lewis baseD. Lewis acid and Lewis base both

The concentration of hydroxyl ion in a solution left after mixing 100 mL of 0.1 M `MgCl_(2)` and 100 mL of 0.2 M NaOH: (`K_(sp)` of `Mg(OH)_(2)=1.2xx10^(-11)`)A. `2.88xx10^(-3)`B. `2.88xx10^(-2)`C. `2.88xx10^(-4)`D. `2.88xx10^(-5)`

The concentration of hydroxyl ion in a solution left after mixing 100 mL of `0.1M MgCl_(2)` and 100 mL of `0.2M NaOH` is: `(K_(SP)of Mg(OH)_(2)=1.2xx10^(-11))`A. `2.8xx10^(-3)`B. `2.8xx10^(-2)`C. `2.8xx10^(-4)`D. `2.8xx10^(-5)`

At `18^(@)C` aniline and acetic acid have dissociation constants `5xx10^(-10)` and `1.8xx10^(-5)` respectively. An aqueous solution of anilium acetate is hydrolysed to the extent of `x%` under equilibrium, what is pH of the solution ? `(K_(w)=10^(-14))`

Solubility of AgCl in pure water is `10^(-5)` mol/litre at `25^(@)`C then, calculate solubility of AgCl in 0.1 M aqueous solution of KCI at `25^(@)` CA. `10^(-9)` mol/litreB. `10^(-5)` mol/litreC. `10^(-7)` mol/litreD. `10^(-4)` mol/litre

The first and second dissociation constants of an acid `H_(2)A` are `1 xx 10^(-5)` and `5 xx 10^(-10)` respectively. The overall dissociation constant of the acid will beA. `9.6 xx 10^(-6)`B. `1.4 xx 10^(-1)`C. `1.2 xx 10^(-6)`D. `1.3 xx 10^(-8)`

The first and second dissociation constant of an acid `H_(2)A` are `1.0xx10^(-5)` and `5.0xx10^(-10)` repectively. The overall dissociation constant of the acid will beA. `5.0xx10^(-15)`B. `0.2xx10^(5)`C. `5.0xx10^(-5)`D. `5.0xx10^(15)`

The first and second dissociation constant of an acid `H_(2)A` are `1.0xx10^(-5)` and `5.0xx10^(-10)` repectively. The overall dissociation constant of the acid will beA. `5.0xx10^(-5)`B. `5.0xx10^(15)`C. `5.0xx10^(-15)`D. `0.2xx10^(5)`

The solubility product of AgCl under standard conditions of temperature is given byA. `1.6 xx 10^(-5)`B. `1.5 xx 10^(-8)`C. `3.2 xx 10^(-10)`D. `1.5 xx 10^(-10)`

The first and second dissociation constants of an acid `H_(2)A` are `1.0 xx 10^(-5)` and `5.0 xx 10^(-10)` respectively. The overall dissociation constant of the acid will beA. `5.0 xx 10^(-5)`B. `5.0 xx 10^(15)`C. `5.0 xx 10^(-15)`D. `0.0 xx 10^(-5)`

In which of the following solution, solubility of AgCl will be maximum?A. 0.1 M `AgNO_(3)`B. WaterC. 0.1 M `-NH_(3)(aq)`D. 0.1 M NaCl

Solid `Ba(NO_(3))_(2)` is gradually dissolved in a `1.0 xx 10^(-4) M Na_(2)CO_(3)` solution. At what concentration of `Ba^(2+)` will a precipitate begin to form (`K_(sp)` for `BaCO_(3) = 5.1 xx 10^(-9)`)A. `4.1 xx 10^(-5) M`B. `5.1 xx 10^(-5) M`C. `8.1 xx 10^(-8) M`D. `8.1 xx 10^(-7) M`

What is the `[HCOO^-]` in the solution that contains 0.015 M HCOOH and 0.02 M HCl ? `K_a(HCOOH)=1.8xx10^(-4)`A. `1.8xx10^(-4)`B. `1.35xx10^(-4)`C. `1.8xx10^(-2)`D. `8xx10^(-3)`

A 0.015 M solution of a weak acid has a pH of 3.52 . What is the value of the `K_a` for this acid?A. `2.0xx10^(-2)`B. `6.2xx10^(-6)`C. `9.1xx10^(-8)`D. `1.4xx10^(-9)`

Calculate the pH of a 0.10 N solution of `H_2CO_3` A. 3.68B. 5.76C. 7.36D. 9.34

A 0.10 M solution of weak acid is 5.75% ionized. What is the `K_a` value for this acid ?A. `3.3xx10^(-3)`B. `3.3xx10^(-4)`C. `4.2xx10^(-5)`D. `3.3xx10^(-5)`

Calculate the change in pH when a 0.1 M solution of `CH_3COOH` in water at `25^@C` is diluted to a final concentration of 0.01 M . `[K_a=1.85xx10^(-5)]`A. `+0.5`B. `+0.4`C. `+0.47`D. `+0.6`

What is the aq. Ammonia concentration of a solution prepared by dissolving 0.15 mole of `NH_4^+CH_3COO^-` in 1 L `H_2O` ? `[K_(a(CH_3COOH))=1.8xx10^(-5)=K_(b(NH_3))]`A. `8.3xx10^(-4)`B. 0.15C. `6.4xx10^(-4)`D. `3.8xx10^(-4)`

The solubility product of AgCl is `1.8xx10^(-10)`. Precipitation of AgCl will occur only when equal volumes of solutions of :A. `10^(-8)MAg^(+)and10^(-8)MCl^(-)`ionsB. `10^(-3)MAg^(+)and10^(-3)MCl^(-)`ionsC. `10^(-6)MAg^(+)and10^(6)MCl^(-)`ionsD. `10^(-10)MAg^(+)and10^(-10)MCl^(-)`ions

Both conc. Being 1 M, the factor by which the percent hydrolysis of `CH_3COO^-` is greater in `CH_3COONH_4` than in `CH_3COONa` is : `[K_a(CH_3COOH)=K_b(NH_4OH)=1.8xx10^(-5)]`A. 236B. 136C. 36D. 3

`2g` of `NaOH` per `250 mL` of solution is added to a buffer solution of buffer capacity `0.2`. Then the change in pH isA. `0.5`B. `1`C. `1.5`D. `2.0`

Buffer capacity of a buffer solution is `x`, the volume of `1 M NaOH` added to `100 mL` of this solution if the change of `pH` by `1` isA. 0.1 x mLB. 10 x mLC. 100 x mLD. x mL

The volume of water that must be added to `100 ml` of `NaOH` solution to change its pH value from `12.6990` to `12` isA. `500 mL`B. `400 mL`C. `100 mL`D. `200 mL`

The solubility product of `BaCrO_(4)` is `2.4xx10^(-10)M^(2)`. The maximum concentration of `Ba(N0_(3))_(2)` possible without precipitation in a `6xx10^(-4)` M `K_(2)CrO_(4)` solution is :A. `4xx10^(-7) M`B. `1.2 xx 10^(-10)` MC. `6xx10^(-4)` MD. `3xx10^(-4)M`.