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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.
| 801. |
`K_(a)`and `K_(b)` are the dissociation constants of weak acid and weak base and `K_(w)` is the ions product of water. Match the pH stated in Column II with the solutions listed in Column I at `25^(@)C`. |
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Answer» Correct Answer - `A rarr q; B rarr C rarr p; D rarr p, s` (A) `KCN + H_(2)O rarr KOH + HCN` (weak acid) `pH = (1)/(2) (pK_(w) + pK_(a) + log[CN^(-)])` `= (1)/(2)[14 + pK_(a) + log 0.1] = 6.5 + (1)/(2)pK_(a)` (B) `C_(6)H_(5)NH_(3)Cl + H_(2)O rarr C_(6)H_(5)NH_(3)OH + HCl` (Cataionic hydrolysis) `pH = (1)/(2)(pK_(a) - pK_(b) - log[C_(6)H_(5)NH_(3)^(+)])` `= (1)/(2)[14 + pK_(b) - log 0.1] = 7.5 + (1)/(2)pK_(b)` (C) KCl - Salt of strong acid (HCl) and strong base (KOH), hence no salt hydrolysis, pH = 7 (D) `CH_(3)COO^(-) + NH_(4)^(+) + H_(2)O rarr CH_(3)COOH + NH_(4)OH` `pH = (1)/(2)(pK_(w)+pK_(b) -pK_(b)) = (1)/(2)[14+pK_(a) - pK_(b)] = 7` (`pK_(a) = pK_(b)` in this case). |
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| 802. |
The pH of a solution is the negative logarithm to the base 10 of its hydrogen ion concentration inA. Moles per litreB. Millimoles per litreC. Micromoles per litreD. Nanomoles per litre |
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Answer» Correct Answer - A `[H^(+)]` is in moles per litre. |
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| 803. |
Aqueous solution of the detertgents areA. NeutralB. AcidicC. BasicD. Amphoteric |
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Answer» Correct Answer - C Detergent is a salt of weak acid and strong base |
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| 804. |
If a salt of weak acid or base is added to a solution of its acid or base respectively, then:A. dissociation of acid or base is diminishedB. pH of the solution in case of acid increase and in case of case decreasesC. mixing of two leads for common ion effectD. none of these |
| Answer» Correct Answer - A::B::C | |
| 805. |
`100 ml` of sample is removed from an aqueous solution saturated with `CaSO_(4)` at `25^(0)C`. The water is completely evaported from the sample and deposit of `0.24 g` of `CaSO_(4)` is obtained. The `Ksp` of `CaSO_(4)` at `25^(0)C` will be.A. `3.115 xx 10^(-4)`B. `3.115 xx 10^(-5)`C. `3.115 xx 10^(-6)`D. `3.115 xx 10^(-3)` |
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Answer» Correct Answer - A `CaSO_(4(s))hArrCa_((aq))^(+2) + SO_(4(aq))^(2-), Ksp = ?` According to the given data, solubility of `CaSO_(4)` is `0.24 g` per `100 ml` `[Ca^(+2)] = [SO_(4)^(2-)] = 0.01765 M` `Ksp = [Ca^(+2)] [SO_(4)^(2-)]` |
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| 806. |
Aqueous solution of salt of strong base and weak acidA. Undergoes anionic hydrolysisB. Is basic in natureC. Has `P^(H)` greater than `7`D. All the above |
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Answer» Correct Answer - D Salt of `W.A + W.B` under goes Animoic hydrolysis, basic in nature and `p^(H)` is greater than 7 |
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| 807. |
A salt of weak acid and weak base undergoesA. Only cationic hydrolysisB. Only anionic hydrolysisC. Both cationic and anionic hydrolysisD. Neither cationic nor anionic hydrolysis |
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Answer» Correct Answer - C Salt of `W.A + W.B` under goes both anionic and cationic hydrolysis |
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| 808. |
The sulphide ion concentration `[S^(2-)]` in saturated `H_(2)S` solution is `1xx10^(-22)`. Which of the following sulphides should be quantitavely precipitated by `H_(2)S` in the presence of dil. `HCl`? `{:("Sulphide",,"Solubility Product"),((I),,1.4xx10^(16)),((II),,1.2xx10^(-22)),((III),,8.2xx10^(-46)),((IV),,5.0xx10^(-34)):}`A. I,IIB. III,IVC. II,III,IVD. Only I |
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Answer» Correct Answer - B Basis radials of group II & IV are precipitated by `H_(2)S` in the form of their sulhphides. Iind group in acidic medium & IV group in acidic medium & IV group in alkaline medium. They precipitate when ionic product increases than solubility product. |
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| 809. |
The wight of `HCl` present in one litre of solution if `pH of the solution is oneA. `3.65 g`B. `36.5 g`C. `0.365 g`D. `0.0365 g` |
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Answer» Correct Answer - A `P^(H) = 1` `:. [H^(+)] = 10^(-PH) = 10^(-1) M` `W = N xx EW xx V ("lit")` `= 0.1 xx 36.5 xx 1` `= 3.65` |
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| 810. |
Which of the following is non-electrolyteA. NaClB. `CaCl_(2)`C. `C_(12)H_(22)O_(11)`D. `CH_(3)COOH` |
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Answer» Correct Answer - C `C_(12)H_(22)O_(11)` is a sugar and non-electrolyte |
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| 811. |
Statement: A micture of the solution of a weak acid and its conjugates base acts as a good buffer. Explanation: The ratio of the conjugates base acid in the mixture does not change substantially when small amount of acids or alkalines are added to the buffer.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. |
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Answer» Correct Answer - D These are facts. |
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| 812. |
Statement: For a weak electrolyte, the plot of molar conductivity `(Lambda_(m))` against `sqrt(C )(C` is concentration in mol `litre^(-1))` is nearly linear. Explanation: The molar conductivity at infinite dilution `(Lambda_(m))` for an electrolyte can be considered equal to the sum of the limiting molar conductivities of the individual ions.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. |
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Answer» Correct Answer - B For a strong electrolyte, the plot of molar conductivity `(Lambda_(m))` against `sqrt(C )` is nearly liner. |
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| 813. |
Ionisation depends uponA. PressureB. VolumeC. DilutionD. None of these |
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Answer» Correct Answer - C Ionisation depends upon dilution, when dilution increases then ionisation is also increased. |
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| 814. |
How many moles of `Ca(OH)_(2)` must be dissolved to produce 250 mL of an aqueous solution of pH `10.65`, assuming completer dissociation ? |
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Answer» Let a mole of `Ca(OH)_(2)` be dissolved in 250 mL. solution to have `pH= 10.65` `[Ca(OH)_(2)]=(axx1000)/(250)=4a M` `{:(("Complete ionization"), Ca(OH)_(2)rarr,Ca^(2+)+,2OH^(-)),("Intial moles", 4a,0,0),("Final moleas", 0,4a, 2xx4aM):}` `:. [OH^(-)]= 8aM` `[H^(+)]=(10^(-14))/(8a)` `.: pH= -llog[H^(+)]` `:. 10.65= -log(10^(-14))/(8a)` `(10^(-14))/(8a)=2.238xx10^(-11)` `:. a=5.85xx10^(-5)`mole |
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| 815. |
Which one of the following salt is most acidic in waterA. `NiCl_(2)`B. `BeCl_(2)`C. `FeCl_(3)`D. `AlCl_(3)` |
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Answer» Correct Answer - D `AlCl_(3)` on hydrolysis give weak base and strong acid among all. |
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| 816. |
The degree of dissociation of water at `25^(@)C` is `1.9 xx 10^(-7)%` and density of `1.0 gm^(-3)`. The ionic constant for water isA. `1.0 xx 10^(-14)`B. `2.0 xx 10^(-16)`C. `1.0 xx 10^(-16)`D. `1.0 xx 10^(-8)` |
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Answer» Correct Answer - B For water `K_(w) = ([H^(+)][OH^(-)])/([H_(2)O]), K = 1.9 xx 10^(-9), C = (100)/(18)` `K = 1.9 xx 10^(-9) xx 1.9 xx10^(-9) xx 1000/18` `= 2.0 xx 10^(-16)` |
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| 817. |
Degree of dissociation of `0.1 N CH_(3)COOH` is (Dissociation constant `= 1 xx 10^(-5)`)A. `10^(-5)`B. `10^(-4)`C. `10^(-3)`D. `10^(-2)` |
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Answer» Correct Answer - D Degree of dissociation `alpha = ?` Normality of solution `= 0.1 N = (1)/(10) N` Dissociation constant `K = 1 xx 10^(-5)` `K = C alpha^(2), alpha = sqrt((K)/(C)) = sqrt((1 xx 10^(-5))/(0.1)) , alpha = 1 xx 10^(-2)`. |
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| 818. |
`K_(a)` for `HA` is `4.9 xx 10^(-8)`. After making the necessary approximation, calculate for its decimolar solution, a. `%` dissociation b. `overset(Θ)OH` concentration c. `pH`A. `3.5`B. `4.2`C. `5.3`D. `2.5` |
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Answer» Correct Answer - B `p^(H) = 1/2(p^(ka) - logc)` `= 1/2(7.3 - "log" 0.1)` `= 1/2(8.3), = 4.15, =4.2` |
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| 819. |
If the pH of a solution is 4.0 a `25^(@)C`, its pOH would be `(K_(w) = 10^(-14))`A. `4.0`B. `6.0`C. `8.0`D. `10.0` |
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Answer» Correct Answer - D pH + pOH = 14, pH = 4.0 pOH = 14 - pH , pOH = 14 - 4.0 = 10.0. |
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| 820. |
If pOH of a solution is 6.0, then its pH will beA. 6B. 10C. 8D. 14 |
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Answer» Correct Answer - C pH+ pOH = 14, pH = 14 - pOH , pH = 14 - 6 = 8. |
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| 821. |
At `25^(@)C`, pH of a `0.01 M` solution of a monobasic acid `[HA]` is 4. The correct statement(s) regarding HA and its given solution is (are)A. HA is a weak acidB. The ionization constant `(K_(a))` of acid is approximately `10^(-6)` at `25^(@)C`.C. Increasing the temperature of solution would cause the `pH` to decrease.D. Addition of `0.1 M HCl` solution would lower pH by increasing degree of ionization |
| Answer» Correct Answer - A::B::C | |
| 822. |
If `H^(+)` ion concentration of a solution is increased by 10 times its pH will beA. Increase by oneB. Remains unchangedC. Decrease by oneD. Increase by 10 |
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Answer» Correct Answer - C When concentration of `[H^(+)]` increased then the value of pH is decreases. `pH = log.(1)/([H^(+)])` |
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| 823. |
`K_(sp)` of salt `A_(3)B_(2)`, for solubility `x mol L^(-1)`, isA. `36 x`B. `72x^(6)`C. `108x^(5)`D. `108x^(6)` |
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Answer» Correct Answer - C `A_(3)B_(2) hArr underset(3x)3overset(+2)(A) + underset(2x)2overset(-3)(B)_(["for solubility of" `A_(3)B_(2)` as xM])` `K_(sp) = [A]^(3) xx [B]^(2) = [3x]^(3) xx [2x]^(2) = 108 x^(5)` |
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| 824. |
The increasing order of basic strength of `CI^(-), CO_(3)^(2-), CH_(3)COO^(-), OH^(-)and F^(-)` is:A. `CI^(-)ltF^(-)ltCH_(3)COO^(-)ltCO_(3)^(2-)ltOH^(-)`B. `CI^(-)ltF^(-)ltCO_(3)^(2-)ltCH_(3)COO^(-)ltOH^(-)`C. `CH_(3)COO^(-)ltCI^(-)ltF^(-)ltCO_(3)^(2-)ltOH^(-)`D. none of these |
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Answer» Correct Answer - A More is the acid character lesser is basic character of conjugate base. |
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| 825. |
The `pH` of a solution increased from `3` to `6`. Its `[H^(o+)]` will beA. reduced to halfB. doubledC. reduced by 1000 timesD. increased by 1000 times |
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Answer» Correct Answer - C When `pH=3, [H^(+)]=10^(-3)` and when pH `=6 ,[H^(+)]=10^(-6)` Thus concentration is reduced by 1000 times. |
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| 826. |
Which buffer solution out of the following will have `pH gt 7`?A. `CH_(3)COOH + CH_(3)COONa`B. `HCOOH +HCOOK`C. `CH_(3)COONH_(4)`D. `NH_(4)OH + NH_(4)Cl` |
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Answer» Correct Answer - D Weak base `+` salt of strong acid . |
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| 827. |
The conjugate acid of `S_(2)O_(8)^(2-)` isA. `H_(2)S_(2)O_(8)`B. `H_(2)SO_(4)`C. `HSO_(4)^(-)`D. `HS_(2)O_(8)^(-)` |
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Answer» Correct Answer - D `underset("conjugate acid")(HS_(2)O_(8)^(-)) hArr underset("conjugate base")(H^(+) + S_(2)O_(8)^(--))`. |
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| 828. |
The conjugated acid of `O^(-2)` ions isA. `OH^(-)`B. `OH^(+)`C. `H_(2)O`D. `HO_(2)^(-)` |
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Answer» Correct Answer - A If addition of proton to species, the resulting species is called conjugate acid `O^(-2) overset(H^(+)) rarr OH^(-)` |
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| 829. |
The species among the following, which can act as an acid and a base isA. `HSO_(4)^(-)`B. `SO_(4)^(2-)`C. `H_(3)O^(+)`D. `Cl^(-)` |
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Answer» Correct Answer - A The species which cab accept as well as donate `H^(+)` can act both as an acid and a base. `underset("base")(HSO_(4)^(-)) + H^(+) hArr H_(2)SO_(4)`. `underset("acid")(HSO_(4)^(-) hArr SO_(4)^(2-) + H^(+)`. |
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| 830. |
In the reaction `HCl + H_(2)O hArr H_(3)O^(+) + Cl^(-)`A. `H_(2)O` is the conjugate base of HCl acidB. `Cl^(-)` is the conjugate base of HCl acidC. `Cl^(-)` is the conjugate acid of `H_(2)O` acidD. `H_(3)O^(+)` is the conjugate base of HCl |
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Answer» Correct Answer - B HCl is a strong acid and their conjugate base is a very weak base. |
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| 831. |
A basic buffer contains equal concentration of base and its salt. The dissociation constant of base is `10^(-6)`. Then the `pH` of the buffer solution isA. `9`B. `8`C. `5`D. `6` |
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Answer» Correct Answer - B `P^(H) = 14 - P^(Kb) - "log" (["Salt"])/(["Base"])` |
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| 832. |
The condition for minimum change in pH for a buffer solution isA. Isoelectronic species are addedB. Conjugate acid or base is addedC. `pH = pK_(a)`D. None of these |
| Answer» Correct Answer - C | |
| 833. |
An acidic buffer contains `0.06 M` salt and `0.02 M` acid. The dissociation constant of acid is `10^(-4)`. The `P^(H)` of the buffer solution isA. `4`B. `10`C. `4.48`D. `9.52` |
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Answer» Correct Answer - C `pH = pKa + "log" ([s])/([A])` |
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| 834. |
The following is a Lewis acidA. `O_(2)`B. `N_(2)`C. `F`D. `O` |
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Answer» Correct Answer - D Sexet electron is lewis acid |
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| 835. |
The salt that forms neutral solution in water isA. `NH_(4)Cl`B. NaClC. `Na_(2)CO_(3)`D. `K_(3)BO_(3)` |
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Answer» Correct Answer - B Because it is a salt of strong acid and strong base. `H_((aq))^(+) + Cl_((aq))^(-) + Na_((aq))^(+) + OH_((aq))^(-) hArr H_(2)O_((I)) + Na_((aq))^(+) + Cl^(-)` |
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| 836. |
An example of a Lewis acid isA. NaClB. `MgCl_(2)`C. `AlCl_(3)`D. `SnCl_(4)` |
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Answer» Correct Answer - C::D `AlCl_(3)` has incomplete octet and `SnCl_(4)` has empty d-orbital. `AlCl_(3)` and `SnCl_(4)` both are Lewis acid. |
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| 837. |
Which of the anhydrous salts when come in contact with water turns blueA. Ferrous sulphateB. Copper sulphateC. Znic sulphateD. Cobalt sulphate |
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Answer» Correct Answer - B Cu(II) complexes are blue. The four water molecules are attached with secondary valencies of the metal atom e.g. `[Cu(H_(2)O)_(4)] SO_(4).H_(2)O`. |
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| 838. |
The strongest conjugate base results fromA. Fromic acidB. Benzoic acidC. Acetic AcidD. Acetylene |
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Answer» Correct Answer - D Weak acid conjugate base is the strongest base |
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| 839. |
`K_(a)` for the reaction, `Fe^(3+)(aq)+H_(2)O(l)hArr Fe(OH)^(2+)(aq) +H_(3)O^(o+)(aq)` is `6.5 xx 10^(-3)`, what is the maximum `pH` value which could be used so that at least `80%` of the total iron `(III)` in a dilute solution exsists as `Fe^(3+)`?A. `2.0`B. `2.4`C. `2.8`D. `1.6` |
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Answer» Correct Answer - D Take `h = 20% rArr [H^(+)] = (K_(a)(1-h))/(h)` |
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| 840. |
Which of the following is not lewis baseA. `NH_(3)`B. `PH_(3)`C. `(CH_(3))_(3)N`D. `HN_(3)` |
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Answer» Correct Answer - D Hydrazoic acid `(HN_(3))` is a Lewis acid. |
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| 841. |
`ZnCl_(2)` undeoes hydrolysis `ZnCl_(2) + H_(2)O hArr Zn(OH)_(2) + 2HCl`. The overall `K_(b)` for `Zn(OH)_(2)` is `2.5 xx 10^(-12)` at `25^(0)C`. The degree of hydrolysis of `0.001 M ZnCl_(2)` solution isxA. `(sqrt((K_(H))/(10^(-3))))^(1//2)`B. `(K_(H))/(10^(-6))`C. `[(K)/(4 xx 10^(-6))]^(1//3)`D. `[(K_(H))/(8 xx 10^(-6))]^(1//4)` |
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Answer» Correct Answer - C `{:(Zn^(2+)+,2H_(2)Orarr,Zn(OH)_(2),+2H^(+)),(1,,0,0),((1-h)C,,hC,2hC):}` `K_(H) = 4C^(2)h^(2),h = ((K_(H))/(4 xx 10^(-6)))^(1//3)` |
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| 842. |
The following is neither a Lewis acid nor a Lewos base.A. `H_(2)O`B. `O`C. `H_(2)SO_(4)`D. `H_(3)O^(+)` |
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Answer» Correct Answer - D `H_(3)O^(+)` is neither lonepair aceptor nor lonepair donor |
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| 843. |
Which of the following is neither a Lewis acid nor a LewisA. `HSO_(4)^(-)`B. `ZnCl_(2)`C. `NH_(4)^(+)`D. `CH_(3)^(+)` |
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Answer» Correct Answer - C `NH_(4)^(+)` is neithr lonpair aceptor nor lonepair donor |
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| 844. |
Which of the following solutions will have `pH = 10` at `298 K`?A. `1 xx 10^(-10)M HCI` solutionB. `1 xx 10^(-4)M NaOH` solutionC. `1xx10^(-10)M NaOH` solutionD. Both (a) and (b) |
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Answer» Correct Answer - B a. `10^(-10)M HCI`, since `[H_(3)O^(o+)] lt 10^(-6)M`, so `pH` is less than `7`, because of the acidic solution. b. `10^(-4) M NaOH`, since `[overset(Theta)OH] = 10^(-4)M , pOH =4, pH = 10`. c. `10^(-10)M NaOH`, Since `[overset(Theta)OH] lt 10^(-6)M, pH` is slighly more tan `7` because of the basic solution. |
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| 845. |
An acid `HA` is `40%` dissociated in an aqueous solution. The hydronium ion concentration of its `0.2M` solution would beA. `0.08 M`B. `0.4M`C. `0.2 M`D. None |
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Answer» Correct Answer - A `{:(,HA+,H_(2)OhArr,H_(3)O^(o+)+,A^(Theta)),("Initial",1,0,0,0),("Final",C(1-alpha),,Calpha,Calpha):}` `[H_(3)O^(o+)] = Calpha (alpha = (40)/(100)=0.4)` `= 0.2 xx 0.4 (C = 0.2M)` `= 0.08`. |
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| 846. |
`pH` for the solution of salt undergoing anionic hydrolysis (say `CH_(3)COONa`) is given by:A. `pH=1/2[pK_w+pK_a+log C]`B. `pH=1/2[pK_w+pK_a-log C]`C. `pH=1/2[pK_w+pK_b-log C]`D. none of the above |
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Answer» Correct Answer - a |
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| 847. |
The indicator used in the titration of iodine against sodium thiosulphate isA. StarchB. `K_(3)Fe(CN)_(6)`C. `K_(2)CrO_(4)`D. Potassium |
| Answer» Correct Answer - A | |
| 848. |
Which halide of nitrogen is least basicA. `NBr_(3)`B. `NI_(3)`C. `NCl_(3)`D. `NF_(3)` |
| Answer» Correct Answer - B | |
| 849. |
Which one is the weakest acidA. `HNO_(3)`B. `HClO_(4)`C. `H_(2)SO_(4)`D. HBr |
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Answer» Correct Answer - A `underset("Acidic character decreases")(HClO_(4) gt H_(2)SO_(4) gt HBr gt HNO_(3))` |
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| 850. |
A solution of `Cu(NO_(3))_(2)` in water is acidic due toA. ionisationB. acidic impuritiesC. hydrolysisD. dissociation |
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Answer» Correct Answer - C `Cu^(+2) +2NO_(3)^(-) +2h_(2)O hArr Cu(OH)_(2)+2H^(+)+2NO_(3)^(-)` |
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