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.
| 651. |
What are pH and pOH values of a neutral solution at a temperature at which `K_(w)=10^(-13)`? |
| Answer» `pK_(w)=pH + pOH `. But `pK_(w)=13`. Also, for neutral solution, pH = pOH . Hence, pH = pOH = 6.5. | |
| 652. |
What is pOH? What is its value for neutral water at 25°C? |
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Answer» pOH is a measure of hydroxide ion (OH-) concentration. pOH = −log10[OH−] For neutral water, the value of pOH is 7 at 25°C. |
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| 653. |
What is the relationship between pH and pOH ? |
| Answer» `pH + pOH = pK_(w) = 14`. | |
| 654. |
The value of the equilibrium constant is less than zero. What does it indicate ? |
| Answer» This indicates that the forward reaction has proceeded only to a small extent before the equilibrium is attained. | |
| 655. |
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, `K_(sp)=6.3xx10^(-18))`. |
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Answer» Correct Answer - The highest molarity for the solution is `2.5 xx 10^(-9) M` Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., `(x)/(2)`. `:. [FeSO_(4)] = [Na_(2)S] = x/2 M` Then, `[Fe^(2+)] = [FeSO_(4)] = x/2 M` Also, `[S^(2-)] = [Na_(2)S] = (x)/(2) M` `FeS_((s)) harr Fe_((aq))^(2+) + S_((aq))^(2-)` `K_(sp) = [Fe^(2+)] [S^(2-)]` `6.3 xx 10^(-18) = ((x)/(2)) ((x)/(2)) ` `(x^(2))/(4) = 6.3 xx 10^(-18)` `rArr x = 5.02 xx 10^(-9)` If the concentrations of both solutions are equal to or less than `5.02 xx 10^(-9) M`, then there will be no precipitation of iron sulphide. |
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| 656. |
Write the demerits of Brownsted-Lowry theory. |
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Answer» It does not explained why CaO, NH3 acts as base and BF3 and CO2 acts as acid. |
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| 657. |
What is the relationship between pH and pOH? |
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Answer» pH + pOH = pKw = 14. |
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| 658. |
The ionization constant at 298 K is 1.8 x 10-4 . Calculate the ionization constant of the corresponding conjugate base. |
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Answer» Ka x Kb =Kw Kb= Kw/Ka =1 x10-14/1.8x10-8 =5.55x10-7 |
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| 659. |
Partial pressure of `O_(2)` in the reaction `2Ag_(2)O(s) hArr 4Ag(s)+O_(2)(g)` isA. `K_(rho)`B. `sqrt(Krho)`C. `3sqrt(K_(rho)`D. `2K_(rho)` |
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Answer» Correct Answer - A |
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| 660. |
At `448^(@)` the equilibrium constant `(K_(c))` for the reaction , `H_(2) (g) + I_(2) (g) hArr 2 HI(g) , "is " 50*5.` Predict the direction in which the reaction will proceed to reach equilibrium at `448^(@)C, " if we start with " 2*0xx 10^(-2) " mol of " HI,1*0 xx 10^(-2) " mol of " H_(2) and 3*0 xx 10^(-2) "mol of " I_(2) " in a " 2*0 " L container"` |
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Answer» Correct Answer - Forward direction Q comes out to be `1*3` which is less than K . |
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| 661. |
A reaction mixture containing `N_(2) " at " 0*50 " atm" , "at " 0*05 " atm" NH_(3) " and " 3*0 " atm of hydrogen is heated to " 450^(@)C.` In which direction the reaction `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " will go if " K_(p)" is" 4*28 xx 10^(-5)`? |
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Answer» Correct Answer - Backward direction Concentration quotient (Q) = `(p_(NH_(3))^(2))/(p_(N_(2)) xx p_(H_(2))^(3)) =((0*50)^(2))/(0*5 xx (3*0)^(3))=0*055` As `Q gt gt K_(p),` reaction will go in the backward direction. |
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| 662. |
Which of the following is an example of homogeneous equilibrium ?A. `2SO_(2(g))+O_(2(g))hArr2SO_(3(g))`B. `C_((s))+H_(2)O_((g))hArrCO_((g))+H_(2(g))`C. `CaCO_(3(s))hArrCaO_((s))+CO_(2(g))`D. `NH_(4)NS_((s))hArrNH_(3(g))+H_(2)S_((g))` |
| Answer» Correct Answer - A | |
| 663. |
At equilibrium of the reaction, ` N_(2)O_(4) (g) hArr 2 NO_(2) (g)`, the obsserved molecular weight of `N_(2)O_(4)` is 80 g `" mol "^(-1)` at 350 K. The percentage of dissociation of `N_(2)O_(4)` (g) at 350 K isA. 0.1B. 0.15C. 0.2D. 0.18 |
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Answer» Correct Answer - B For the given dissociation reaction , `alpha =(D-d)/d = (M_(t)-M_(o))/(M_(o))=(92 xx80)/80 xx 100 = 15%` |
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| 664. |
Calculate the degree of hydrolysis of 0.04 M solution of `NH_(4) CI" of " pH =5.28` |
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Answer» Correct Answer - `1.312 xx 10^(-4)` `pH =5.28 " or " log [(1)/(H^(+))] =5.28` `[(1)/(H^(+))] = " Antilog " 5.28 " or " [H^(+)] = " Antilog " (-5.28)` `[H^(+)] = " Antilog " (bar(6).72) =5.248 xx 10^(-6)` Degree of hydrolysis `(h) =[[H^(+)]]/(C) = (5.248 xx 10 ^(-6))/(0.04) =1.312 xx 10^(-4)` |
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| 665. |
The vapour density of `PCl_(5)` at 473 is found to be` 70*2`. Find the degree of dissociation of `PCl_(5)` at this temperature. |
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Answer» Theoretical vapour density of `PCl_(5)` , `D=(" Mol.mass of " PCl_(5))/2 = (31 + 5 xx 35*5)/2 = (208*5)/2 = 104 * 25` Observed vapour density , d= `70*2 ` `:. "Degree of dissociation " (alpha)=(104*25 - 70*2)/(70*2)=0*485` |
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| 666. |
Which of the following statement is correct for a reversible process in a state of equilibrium ?A. ` Delta G^(@) = - 2*30 RT log K `B. ` DeltaG^(@) = 2*30 RT log K`C. `Delta G- -2*30 RT log K`D. `Delta G = 2.30 RT log K` |
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Answer» Correct Answer - A For a reversible process in a state of equilibrium , `Delta G^(@)=-2.30" RT "log K. ` |
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| 667. |
Consider the following equilibrium in a closed container `N_(2)O_(4)(g) hArr 2NO_(2)(g)` At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statement holds true regarding the equilibrium constant `(K_(p))` and the degree of dissociation `(alpha)`?A. Neither `K_(p) " nor " alpha` changesB. Both `K_(p) " and "alpha` changeC. `K_(p)` changes but `alpha` does not changeD. `K_(p)` does not change order of solubility of : |
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Answer» Correct Answer - D When volume is dicreased. Pressure increases and this will favour bacward reaction. Thus degree of dissociation decreases but `K_(p).` does not change since temp. is constat. |
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| 668. |
Equilibrium concentrations of A, B and C in a reversible reaction `3A+BhArr 2C+D` are `0.03`, `0.01`, and `0.008 mol L^(-1)`. Calculate the initial concentration of A?A. `0.014`B. `0.042`C. `0.084`D. `0.343` |
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Answer» Correct Answer - B `{:(,3A+BhArr2C+D),("Initial (M)"," a b 0 0"),("Change (M)"," -3x -x +2x +x"),("Equilibrium (M)",bar("(a-3x) (b-x) 2x x ")):}` According to the data, `2x=0.008` `x=0.004` `:. (a-3x)=[a-3(0.004)]=0.03` `a=0.03+0.012` `=0.042` Alternative method: According to the stoichiometry of the reaction, 2 mol C consume 3 mol A. 1 mol C consumes `3//2 mol A` `0.008 mol C` consume `3/2xx(0.008)=0.012 mol A` `:.` Initial concentration of `A` `=` Equilibrium concentration `+` Concentration reacted at equilibrium `=0.03+0.012=0.042` |
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| 669. |
Consider the following equilibrium in a closed container : `N_(2)O_(4) (g) hArr 2 NO_(2)(g)` At a fixed temperature , the volume of the reaction container is halved . For this change , which of the following statement , holds true regarding the equilibrium constant `(K_(p))` and degree of dissociation `(alpha)` ?A. neither `K_(p) "nor" alpha` changesB. both `K_(p) and alpha ` changeC. `K_(p) " changes but " alpha ` does not changeD. `K_(p) ` does not change but `alpha` changes |
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Answer» Correct Answer - D `K_(p)` is constant at constant temperature . As volume is halved , pressure will be doubled . Hence, equilibrium will shift in the backward direction , i.e, degree of dissociation decreases . |
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| 670. |
Mention atleast three ways by which the concentration of `SO_(2)(g)` be increased in the following reaction in a state of equilibrium : `2SO_(2) (g) + O_(2) (g) hArr 2SO_(3) (g) +` heat. |
| Answer» Correct Answer - (i) Lowering temperature (ii) Increasing pressure (iii) Increasing concentration of oxygen | |
| 671. |
At half-neutralisation of a weak acid with a strong base, what is the relationship between pH and dissociation constant `(K_(a))` of the weak acid ? |
| Answer» at half- neutralisation, `pH = pK_(a)`. | |
| 672. |
STATEMENT-1: Autoprotolysis constant of water increases with the increase in temperature. STATEMENT-2: When a solution of a weak monobasic acid is titrated wita a strong base, at half neutralization point `pH=pK_(a)+1.` STATEMENT-3: The pH of `10^(-8)` m HCl is 8.A. F F TB. T F FC. F T TD. T T T |
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Answer» Correct Answer - B |
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| 673. |
Autoprotolysis constant of `NH_(3)` isA. `[NH_(4)^(+)][NH_(3)]`B. `[NH_(2)^(-)][NH_(3)]`C. `[NH_(4)^(+)][NH_(2)^(-)]`D. `[NH_(2)^(-)]//[NH_(2)^(-)]` |
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Answer» Correct Answer - C `NH_(3)+NH_(3) hArr NH_(4)^(+)+NH_(2)^(-)` Hence, `K=([NH_(4)^(+)][NH_(2)^(-)])/([NH_(3)]^(2))` `:. [NH_(4)^(+)][NH_(2)^(-)]=K[NH_(3)]^(2)` = Autoprotolysis constant. |
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| 674. |
Calculate the pH of a solution which is `1xx10^(-3)` M with respect to sulphuric acid. |
| Answer» Correct Answer - `2.699` | |
| 675. |
What will be the pH of `1xx10^(-4)` M `H_(2)SO_(4)` solution ?A. 10.4B. 3.7C. 3D. 13 |
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Answer» Correct Answer - B `H_(2)SO_(4)hArr2H^(+)+SO_(4)^(2-)` `[H^(+)]=2xx1xx10^(-4)M` `pH=-log(2xx10^(-4))=3.70` |
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| 676. |
Which concept can justify that `CaO + SO_(3) rarr CaSO_(4) ` is an acid-base reaction ? |
| Answer» Correct Answer - Lewis concept. | |
| 677. |
Calculate the pH value of a solution obtained by mixing 50 mL of 0.2 N HCl solution with 50 mL of 0.1 N NaOH solution. |
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Answer» When equal volumes of HCI and NaOH solutions are mixed , the volume becomes double and the corresponding normality (or molarity is reduced to half. Normality of HCI solution on mixing `=((0.2N) xx(50 mL))/((100 mL)) =0.1N` `[H^(+)]` in solution `=0.1 N= 0.1 M` Normality of NaOH solution on mixing `=((0.1N)xx(50mL))/((100mL)) =0.05 N` `[OH^(-)]` in solution `=0.05 N= 0.05 M` `[H^(+)]` in solution after neutralisation `=0.1 -0.05 =0.05 M` pH of solution `=-log [H^(+)] =- log (0.05) = - log (5xx10^(-2))` `=- (log 5-2 log 10) = (2-log 5) =( 2-0.6989) = 1.301` |
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| 678. |
Among the following, which causes the greatest change in pH on addition to 50 ml of 0.2 M oxalic acid solution?A. Addition of 25 ml of 0.02 M oxalic acidB. Addition of 25 ml of 1 M NaOH solutionC. Addition of 2 ml of 0.02 M `NH_(4)OH` solutionD. Addition of 50 ml of 0.2 M acetic acid solution |
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Answer» Correct Answer - B |
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| 679. |
What will be the `H^(+)` concentration in a solution prepared by mixing 50.0 ml of 0.20 m NaCl, 25 ml of 0.10 M NaOH and 25.0 ml of 0.30 M HCl?A. `0.5` MB. `0.05` MC. `0.02` MD. `0.10` M |
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Answer» Correct Answer - B |
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| 680. |
If pH of a solution is 7, calculate its pOH value. |
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Answer» pH +pOH = 14 ∴ pOH = 14 – pH = 14 – 7 = 7. |
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| 681. |
What happens to the pH if a few drops of acid are added to CH3COONH4 solution? |
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Answer» pH will almost remain constant. |
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| 682. |
The pH of a 10-10M NaOH solution is nearest to(a) 10 (b) 7(c) 4 (d) -10 |
| Answer» The correct answer is (b) 7 | |
| 683. |
What is the structure of H3O+ ion? |
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Answer» Three are three O-H bond pairs and one lone pair on oxygen atom in H3O+ ion so the ion has pyramidal shape. |
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| 684. |
What is the concentration of H3O+ and OH- ions in water at 298K? |
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Answer» [H3O+] = [OH-] = 1 x 10-7 mol-1 |
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| 685. |
A certain sample of beer has a pH of 10. The H3O+ ion concentration of the beer is(a) 1010M (b) 10-2M(c) 10-4M (d) 10-10M |
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Answer» (d) The H3O+ ion concentration of the beer is 10-10M. |
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| 686. |
The vapor pressures of water, acetone, and ethanol at `293K` are `2.34, 12.36`, and `5.85 kPa`, respectively. Which of the following statements is correct?A. Acetone has the lowest boiling point.B. Water has the highest boiling point.C. Water evaporates the least in a sealed container at `293K` before equilibrium is established.D. All of these |
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Answer» Correct Answer - D When the temperature rises, the vapor pressure of a liquid also rises. At boiling point, the vapor pressure becomes equal to the atmospheric pressure. A liquid with lower vapor pressure will require higher temperature to attain its vapor pressure will require higher temperature to attain its vapor pressure equal to the atmospheric pressure. Thus, the liquid with the lowest vapor pressure will have the highest boiling point. |
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| 687. |
Pure water is netural in nature `[H^(+)]=[OH^(-)].` When this condition is disturbed by changing the concentration of `H^(+)or OH^(-),` the natural solution changes to acidic `{[H^(+)]gt[OH^(-)]}` or basic `{[H^(+)]lt[OH^(-)]}.` This change occurs during salt hydrolysis. pH of salt solution can be calculate using the following relation (i) Salt of weak acid and strong base `pH =1/2[pK_(w)+pK_(a)+logC]` (ii) Salt of weak base and strong acid `pH=1/2[pK_(w)-pK_(b)-logC]` (iii) For salt of weak base and strong acid `pH=1/2[pK_(w)+pK_(a)-pK_(b)]` The pH of buffer can be calculated using t he following formula `pH=pK_(a)+log""(["Salt"])/(["Acid"])` `pOH=pK_(b)=log""(["Salt"])/(["Base"])` Answer t he following questions when `pK_(a)=4.7447` `pK_(b)=4.7447` ltb rgt `pK_(w)=14` 1 mole of `CH_(3)COOH` is dessolved in water to from 1 litre aqueous solution. The pH of resulting solution will beA. `9.2253`B. `2.3723`C. 14D. 7 |
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Answer» Correct Answer - B |
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| 688. |
If the concentration of two monobasic acids are same, their relative strength can be compared byA. `(alpha_(1))/(alpha_(2))`B. `(K_(1))/(K_(2))`C. `([H^(+)]_(1))/([H^(+)]_(2))`D. `sqrt((K_(1))/(K_(2)))` |
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Answer» Correct Answer - A::C::D |
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| 689. |
For which of the following reaction `K_(p)=K_(c)` ?A. `NOCI(g) hArr 2NO(g) +CI_(2) (g)`B. `N_(2) (g) +3H_(2) (g) hArr 2NH_(3)(g)`C. `H_(2)(g) +CI_(2) (g) hArr 2HCI(g)`D. `PCI_(3) (g) + CI_(2) (g) hArr PCI_(5) (g)` |
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Answer» Correct Answer - C `K_(p) =K_(c) " when " Deltan =0` `" In " c " choice " Deltan =2 -2 =0` |
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| 690. |
STATEMENT-1 Net reaction can occur only if a system is in equlibrium. STATEMENT-2: All reactin tends to be in a state of equlibrium. STATEMENT-3: At equlibrium, `DeltaG` is zero.A. T T FB. F T TC. T T TD. F T F |
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Answer» Correct Answer - C |
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| 691. |
Which of the following solution will have no effect on Ph on dilution ?A. `0.1 M NH_(4)OH+0.1 M NH_(4)Cl`B. `0.5MH_(2)CO_(3)+0.5M NaHCO_(3)`C. `1M CH_(3)COOHN_(4)`D. `0.1 M CH_(3)COONH_(4)` |
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Answer» Correct Answer - A::B::C |
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| 692. |
Match the Colume I with Column II |
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Answer» Correct Answer - A(q), B(s), C(p, r), D(s) |
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| 693. |
`AlCI_(3)` is the an acid according to which concept ? |
| Answer» Correct Answer - Lewis concept | |
| 694. |
Pure water is netural in nature `[H^(+)]=[OH^(-)].` When this condition is disturbed by changing the concentration of `H^(+)or OH^(-),` the natural solution changes to acidic `{[H^(+)]gt[OH^(-)]}` or basic `{[H^(+)]lt[OH^(-)]}.` This change occurs during salt hydrolysis. pH of salt solution can be calculate using the following relation (i) Salt of weak acid and strong base `pH =1/2[pK_(w)+pK_(a)+logC]` (ii) Salt of weak base and strong acid `pH=1/2[pK_(w)-pK_(b)-logC]` (iii) For salt of weak base and strong acid `pH=1/2[pK_(w)+pK_(a)-pK_(b)]` The pH of buffer can be calculated using t he following formula `pH=pK_(a)+log""(["Salt"])/(["Acid"])` `pOH=pK_(b)=log""(["Salt"])/(["Base"])` Answer t he following questions when `pK_(a)=4.7447` `pK_(b)=4.7447` ltb rgt `pK_(w)=14` When 50 ml of 0.1 m NaOH is added of 50 ml of `0.1 MCH_(3)COOH` solution the pH will beA. `4.7447`B. `9.2553`C. `87218`D. `1.6020` |
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Answer» Correct Answer - C |
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| 695. |
Which of the following is correct about the equlibrium ?A. Cayalyst has no effect on equlibrium stateB. `K_(aq)` changes with temperatueC. Value of `K_(eq)` changes by increasing concentration of equilibriumD. `Delta G=0` |
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Answer» Correct Answer - A::B::D |
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| 696. |
Match the following |
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Answer» Correct Answer - A(p, q, s), B(q, r, s), C(r, s,t ), D( r, s, t ) |
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| 697. |
The pH of the neutralisation point of 0.1 N ammonium hydroxide with 0.1 NHCI isA. 1B. 6C. 7D. 9 |
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Answer» Correct Answer - B `NH_(4)OH +HCI to NH_(4)CI` `NH_(4)CI` is a salt of weak base and strong acid . Hence it gives the acidic solution with `pH lt 7` |
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| 698. |
Pure water is netural in nature `[H^(+)]=[OH^(-)].` When this condition is disturbed by changing the concentration of `H^(+)or OH^(-),` the natural solution changes to acidic `{[H^(+)]gt[OH^(-)]}` or basic `{[H^(+)]lt[OH^(-)]}.` This change occurs during salt hydrolysis. pH of salt solution can be calculate using the following relation (i) Salt of weak acid and strong base `pH =1/2[pK_(w)+pK_(a)+logC]` (ii) Salt of weak base and strong acid `pH=1/2[pK_(w)-pK_(b)-logC]` (iii) For salt of weak base and strong acid `pH=1/2[pK_(w)+pK_(a)-pK_(b)]` The pH of buffer can be calculated using t he following formula `pH=pK_(a)+log""(["Salt"])/(["Acid"])` `pOH=pK_(b)=log""(["Salt"])/(["Base"])` Answer t he following questions when `pK_(a)=4.7447` `pK_(b)=4.75` lt rgt `pK_(w)=14` When 50 ml of 0.1 M `NH_(4)OH` is added to 50 ml of 0.05 M HCl solution, the pH is nearlyA. `1.60`B. `12.40`C. `4.75`D. `9.25` |
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Answer» Correct Answer - D |
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| 699. |
Which of the following statements are correct ?A. The pH of `1.0xx10^(-8)` M solution of HCl is 8B. The conjugate base of `H_(2)PO_(4)^(-2)`C. Auto-protolysis constant of water increases with temperatureD. neutralization point ` pH =((1)/(2))pK_(a)` |
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Answer» Correct Answer - B::C |
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| 700. |
Equal volumes of solution of `pH=6and pH=8` are mixed. What will be the pH of resulting mixture? |
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Answer» Correct Answer - 7 |
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