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QUESTION: 1

Solubility of MX_{2}-type eletrolytes is 0.5 × 10^{–4} mole/lit, then find out K_{sp }of electrolytes [2002]

Solution:

Given s = 0.5 × 10^{–4} moles/lit

QUESTION: 2

Which has the highest value of pH? [2002]

Solution:

Na_{2}CO_{3} is a salt of weak acid H_{2}CO_{3} and strong base NaOH, therefore, its aqueous solution will be basic hence has pH more than 7.

QUESTION: 3

Solution of 0.1 N NH_{4}OH and 0.1 N NH_{4}Cl has pH 9.25. Then find out pK_{b} of NH_{4}OH [2002]

Solution:

but pOH+ pH = 14 or pOH = 14 – pH

14 – 9.25 – 0 = pK_{b }pK_{b} = 4.75

QUESTION: 4

The reaction quotient (Q) for the reaction

is given by

The reaction will proceed from right to left if [2003]

where K_{c} is the equilibrium constant

Solution:

For reaction to proceed from right to left

i.e the reaction will be fast in backward direction i.e r_{b} > r_{f}.

QUESTION: 5

Which one of the following orders of acid strength is correct? [2003]

Solution:

The higher is the tendency to donate proton, stronger is the acid. Thus the correct order is R – COOH > HOH > R – OH > CH ≡ CH depending upon the rate of donation of proton.

QUESTION: 6

Which one of the following compounds is not a protonic acid? [2003]

Solution:

B(OH)_{3 }does not provide H^{+} ions in water instead it accepts OH^{–} ion and hence it is Lewis acid

QUESTION: 7

The solubility product of AgI at 25ºC is 1.0 × 10^{–16} mol^{2 }L^{–2}. The solubiliy of AgI in 10^{–4} N solution of KI at 25ºC is approximately(in mol L^{–1}) [2003]

Solution:

K_{sp} for AgI = 1 × 10^{–16} In solution of KI, I^{–} would be due to the both AgI and KI, 10^{–4} solution KI would provide = 10^{–4} I^{–}

AgI would provide, say = x I^{–} (x is solubility of AgI)

as x is very small ∴ x^{2} can be ignored

∴ 10^{–4 }x = 10^{–16}

QUESTION: 8

The solubility product of a sparingly soluble salt AX_{2} is 3.2 x 10^{-11} . Its solubility ( in moles/litre) is

[2004]

Solution:

For

QUESTION: 9

The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentrations of the conjugate acid (HIn) and base (In^{–}) forms of the indicator by the expression [2004]

Solution:

For an acid-base indicator

Taking negative on both sides

or we can write

QUESTION: 10

H_{2}S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is be cau se [2005]

Solution:

IV^{th }group needs higher S^{2–} i on concentration. In presence of HCl, the dissociation of H_{2}S decreases hence produces less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H_{2}S which is sufficient to precipitate II^{nd} group radicals.

QUESTION: 11

What is the correct relation ship between the pHs of isomolar solutions of sodium oxide (pH_{1}), sodium sulphide (pH_{2}), sodium selenide (pH_{3}) and sodium telluride (pH_{4})? [2 00 5]

Solution:

The solution formed from isomolar solutions of sodium oxide, sodium sulphide, sodium selenide H_{2}O, H_{2}S, H_{2}Se & H_{2}Te respectively.

As the acidic strengh increases from H_{2}O to H_{2}Te thus pH decreases and hence the correct of pHs is pH_{1} > pH_{2} > pH_{3} > pH_{4}.

QUESTION: 12

At 25°C, the dissociation con stant of a base, BOH, is 1. 0 x 10^{-12}. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be [2005]

Solution:

Given K_{b} = 1.0 × 10^{–12}

On calculation, we get, x = 1.0 × 10^{–5}

Now [OH^{–}] =cx

= 0.01 × 10^{–5}

= 1 × 10^{–7}mol L^{–1}

QUESTION: 13

For the reaction [2006]

Which of the following statements is not true ?

Solution:

First option is incorrect as the value of K_{P} given is wrong. It should have been

QUESTION: 14

The hydrogen ion concentration of a 10^{–8} M HCl aqueous solution at 298 K (K_{w }= 10^{–14}) is [2006]

Solution:

For a solution of 10^{–8} M HCl [H^{+}] = 10^{–8} [H^{+}] of water = 10^{–7 }

Total [H^{+}] = 10^{–7} + 10^{–8} = 10 × 10^{–8} + 10^{–8}

10^{–8} (10 + 1) = 11 × 10^{–8}

QUESTION: 15

Which of the following pairs constitutes a buffer?

Solution:

HNO_{2 }is a weak acid and NaNO_{2} is salt of that weak acid and strong base (NaOH).

QUESTION: 16

A weak acid, HA, has a K_{a} of 1.00 × 10^{–5}. If 0.100 mole of this acid dissolved in one litre of water, the percentage of acid dissociated at equilbrium is closest to [2007]

Solution:

Given K_{a} = 1.00×10^{–5}, C= 0.100 mol for a weak electrolyte, degree of dissociation

QUESTION: 17

The following equilibrium constants are given:

The equilibrium constant for the oxidation of NH_{3} by oxygen to give NO is

Solution:

Given,

We have to calculate

For this equation,

Now operate,

QUESTION: 18

Calculate the pOH of a solution at 25°C that contains 1× 10^{–10} M of hydronium ions, i.e. H_{3}O^{+}.

Solution:

Given [H_{3}O^{+}] = 1 × 10^{–10} M at 25º [H_{3}O^{+}] [OH^{–}] = 10^{–14}

∴ pOH = 4

QUESTION: 19

Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H^{+} ion concentration in the mixture ? [2008]

Solution:

[H_{3}O]^{+} for a solution having pH = 3 is given by [H_{3}O]^{+ }= 1×10^{–3 }moles/litre

Similarly for solution having pH = 4, [H_{3}O]^{+} = 1 × 10^{–4} moles/ litre and for pH=5 [H_{3}O^{+}] = 1×10^{–5} moles/ litre Let the volume of each solution in mixture be IL, then total volume of mixture solution L = (1 + 1 + 1) L =3L

Total [H_{3}O]^{+} ion present in mixture solution = (10^{–3} + 10^{–4} + 10^{–5}) moles Then [H_{3}O]^{+ }ion concentration of mixture solution

= 0.00037 M = 3.7 ×10^{–4} M.

QUESTION: 20

If the concentration of OH^{–} ions in the reaction decreased by times, then equilibriumconcentration of Fe^{3+} will increase by : [2008]

Solution:

For this reaction K_{eq}. is given by

If (OH^{–}) is decreased by times then forreaction equilibrium constant to remain constant, we have to increase the concentration of [Fe^{3+}] by a factor of 43 i.e 4× 4 × = 64. Thus option (c) is correct answer.

QUESTION: 21

Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH ? [2008]

Solution:

The highest pH will be recorded by the most basic solution. The basic nature of hydroxides of alkaline earth metals increase as we move from Mg to Ba and thus the solution of BaCl_{2} in water will be most basic and so it will have highest pH.

QUESTION: 22

The dissociation equilibrium of a gas AB_{2} can be represented as : [2008]

The degree of dissociation is ‘x’ and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K_{p} and total pressure P is :

Solution:

For the reaction

= x^{3} [(1–x) can be neglected in denominator

The partial pressure at equilibrium are calculated on the basis of total number of moles at equilibrium.

Total number of moles

= 2 (1–x) + 2x + x = (2 + x)

where P is the total pressure.

Since x is very small so can be neglected in denominator Thus, we get

QUESTION: 23

The values of Kp_{1} and Kp_{2} for the reactions

are in the ratio of 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio :

Solution:

Given reaction are

Let the total pressure for reaction (i) and (ii) be P_{1} and P_{2} respectively, then

After dissociation,

At equilibrium (1–α) α α

[Let 1 mole of X dissociate with α as degree of dissociation ]

Total number of moles = 1– α + α + α = (1+α)

Thus

We have,

Dividing (i) by (ii), we get

i.e. Option (c) is correct answer.

QUESTION: 24

The value of equilibrium constant of the reaction

[2008]

The equilibrium constant of the reaction

Solution:

Given : Equilibrium constant (K_{1}) for the reaction

To find equilibrium constant for the following reaction

For this multiply (i) by 2, we get

[Note: When the equation for an equilibrium is multiplied by a factor, the equilibrium constant must be raised to the power equal to the factor]

Now reverse equation (iii), we get

[Note: For a reversible reaction, the equilibrium constant of the backward reaction is inverse of the equilibrium constant for the forward reaction.] Equation (iv) is the same as the required equation (ii), thus K_{2} for equation (ii) is i.e.

option (b) is correct.

QUESTION: 25

The dissociation constants for acetic acid and HCN at 25°C are 1.5 × 10^{–5} and 4.5 × 10^{–10} respectively. The equilibrium constant for the equilibrium [2009] would be:

Solution:

Given

K_{a1} , = 1.5 × 10^{– 5 .}...(i)

∴ From (i) and (ii), we find that the equilibrium constant (K_{a}) for the reaction ,

QUESTION: 26

Which of the following molecules acts as a Lewis acid ?[2009]

Solution:

(CH_{3})_{3 }B - is an electron deficient, th us behave as a lewis acid.

QUESTION: 27

The ionization constant of ammonium hydroxide is 1.77 × 10^{–5} at 298 K. Hydrolysis constant of ammonium chloride is: [2009]

Solution:

Ammonium chloride is a salt of weak base and strong acid. In this case hydrolysis constant K_{h} can be calculated as

QUESTION: 28

If pH of a saturated solution of Ba (OH)_{2} is 12, the value of its K_{(sp)} is : [2010]

Solution:

pH = 12 or pOH = 2

[∴ Concentration of Ba ^{2+} is half of OH^{-}]

QUESTION: 29

What is [H^{+}] in mol/L of a solution that is 0.20 M in CH_{3}COONa and 0.10 M in CH_{3}COOH?K_{a} for CH_{3}COOH = 1.8 × 10^{-5} . [2010]

Solution:

QUESTION: 30

In which of the following equilibrium K_{c} and K_{p} are not equal? [2010]

Solution:

Δn = 2 – 1 = + 1

∴ K_{c} and K_{p} are not equal.

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