JEE Chemistry Equilibrium MCQs Set 33

Answer: (c) 7.02

Explanation:
1. For a salt of a weak acid and a weak base, \( \text{pH}=7+\tfrac12(pK_a-pK_b) \).
2. This expression does not contain the concentration, so dilution does not change the pH.

Teacher's Note:
1. Salts of weak acid plus weak base: pH is independent of concentration.

Answer: (a) \( \ce{C6H5NH3+} \), \( \ce{CH3COO-} \)

Explanation:
1. The relation applies to salts in which both ions hydrolyse, that is, the cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid.
2. \( \ce{C6H5NH3+} \) (from weak base aniline) and \( \ce{CH3COO-} \) (from weak acid) both hydrolyse.
3. The other pairs contain a spectator ion from a strong acid or base.

Teacher's Note:
1. Both ions must be hydrolysable for the formula to apply.

Answer: (d) \( 10^{7}\left(\frac wM\right)^5 \)

Explanation:
1. Solubility in mol per litre: \( s=\frac{w\times10}{M} \).
2. For \( \ce{Ca3(PO4)2} \), \( K_{sp}=(3s)^3(2s)^2=108\,s^5 \).
3. \( K_{sp}=108\times10^5\left(\frac wM\right)^5\approx10^7\left(\frac wM\right)^5 \).

Teacher's Note:
1. Convert g per 100 mL to mol per litre by multiplying by 10 and dividing by M.
2. Ca3(PO4)2 gives 3 Ca2+ and 2 PO4(3-).

Answer: (b) 0.01%

Explanation:
1. \( K_h=\frac{K_w}{K_a}=\frac{10^{-14}}{10^{-5}}=10^{-9} \).
2. \( h=\sqrt{K_h/C}=\sqrt{10^{-8}}=10^{-4} \), that is 0.01%.

Teacher's Note:
1. Degree of hydrolysis of a weak-acid salt = square root of (Kh / C).

Answer: (b) \( \ce{H2O} \)

Explanation:
1. Of the two acids in the equilibrium, \( \ce{H3O+} \) is the stronger.
2. Its conjugate base, formed by loss of a proton, is \( \ce{H2O} \).

Teacher's Note:
1. Strong acid has a weak conjugate base.
2. The conjugate base differs from the acid by one proton.

Answer: (d) \( 6.02\times10^{7} \)

Explanation:
1. \( [H^+]=10^{-13} \) mol per litre, so 1 mL contains \( 10^{-16} \) mol.
2. Number of ions = \( 10^{-16}\times6.02\times10^{23}=6.02\times10^{7} \).

Teacher's Note:
1. Convert concentration to moles in the given volume first.

Answer: (a) (b), (c) and (d)

Explanation:
1. A buffer needs a weak acid with its conjugate base, or a weak base with its conjugate acid.
2. (b) boric acid with borate, (c) dihydrogen phosphate with hydrogen phosphate and (d) ammonia with ammonium all qualify.
3. (a) NaCl is neutral and NaOH is a strong base, so no buffer forms.

Teacher's Note:
1. Look for a weak acid/base pair, not a strong one.

Answer: (c) \( 4\times10^{-12} \) M

Explanation:
1. [Hsac] = \( \frac{4\times10^{-4}}{0.2}=2\times10^{-3} \) M, and \( [H^+]=10^{-3} \) M.
2. \( [Sac^-]=\frac{K_a[Hsac]}{[H^+]}=\frac{2\times10^{-12}\times2\times10^{-3}}{10^{-3}}=4\times10^{-12} \) M.

Teacher's Note:
1. The common H+ keeps the ionisation tiny, so Hsac stays essentially undissociated.

Answer: (c) \( \ce{NH2CH2COO-} \), \( \ce{NH3+CH2COOH} \)

Explanation:
1. Removing a proton from the zwitterion (from the \( \ce{NH3+} \) group) gives the conjugate base \( \ce{NH2CH2COO-} \).
2. Adding a proton (to the \( \ce{COO-} \) group) gives the conjugate acid \( \ce{NH3+CH2COOH} \).

Teacher's Note:
1. Conjugate base = species minus a proton; conjugate acid = species plus a proton.

Answer: (a) C < B < A

Explanation:
1. Sodium acetate hydrolyses to give a basic solution (pH above 7).
2. The buffer B has pH close to \( pK_a \) (about 4.76).
3. Acetic acid alone is more acidic than the buffer, so C < B < A.

Teacher's Note:
1. Adding the common ion (acetate) to a weak acid raises its pH.

Answer: (d) 1.4%

Explanation:
1. Concentration = 0.05 M.
2. \( \alpha=\sqrt{K_a/C}=\sqrt{10^{-5}/0.05}=\sqrt{2\times10^{-4}}=0.0141 \).
3. Percent dissociation is 1.4%.

Teacher's Note:
1. Divide the moles by the volume first.

Answer: (b) 8.7

Explanation:
1. At equivalence the solution is potassium acetate with concentration 0.05 M (volume doubled).
2. \( pH=7+\tfrac12(pK_a+\log C)=7+\tfrac12(4.70-1.30)=8.7 \).

Teacher's Note:
1. The equivalence point of a weak acid with a strong base is basic.
2. Dilution on mixing halves the salt concentration.

Answer: (b) 3

Explanation:
1. Moles = \( \frac{0.18}{180}=10^{-3} \), so concentration = 0.01 M.
2. \( [H^+]=\sqrt{K_aC}=\sqrt{10^{-4}\times10^{-2}}=10^{-3} \) M.
3. pH = 3.

Teacher's Note:
1. Total mass of two tablets is 0.18 g.
2. Treat it as a weak monoprotic acid.

Answer: (d) \( 1\times10^{-5} \)

Explanation:
1. \( [H^+]=10^{-5.48}=3.3\times10^{-6} \) M.
2. \( K_a=[H^+]\frac{[In^-]}{[HIn]}=3.3\times10^{-6}\times\frac{75}{25}=1\times10^{-5} \).

Teacher's Note:
1. Ratio of ionised to unionised form is 75 : 25.

Answer: (a) \( \ce{(CH3)2NH} \)

Explanation:
1. The conjugate base is obtained by removing one proton.
2. Removing \( \ce{H+} \) from \( \ce{(CH3)2NH2+} \) gives \( \ce{(CH3)2NH} \).

Teacher's Note:
1. Subtract exactly one H+.

Answer: (c) If \( S_1 \) and \( S_2 \) are the molar solubilities of AX and \( \ce{BX2} \), then \( S_1=2(S_2)^{2/3} \)

Explanation:
1. For AX, \( K_{sp}=S_1^2 \); for \( \ce{BX2} \), \( K_{sp}=4S_2^3 \).
2. Equating gives \( S_1=2S_2^{3/2} \), not \( 2S_2^{2/3} \), so the first statement is incorrect.
3. For small Ksp the 1:1 salt is less soluble than the 1:2 salt, H+ removes a weak-acid anion and raises solubility, and heating raises solubility of these salts.

Teacher's Note:
1. Write the Ksp expression for each salt and equate them.
2. The exponent is 3/2, not 2/3.

Answer: (c) 0.2%

Explanation:
1. \( [OH^-]=K_b\frac{[\text{base}]}{[\text{salt}]}=2\times10^{-5}\times1=2\times10^{-5} \) M.
2. \( \alpha=\frac{2\times10^{-5}}{0.01}=2\times10^{-3} \), that is 0.2%.

Teacher's Note:
1. The common ion NH4+ suppresses dissociation of the base.

Answer: (a) \( \ce{BaCrO4} \)

Explanation:
1. The salt with the smallest solubility product precipitates first at equal ion concentrations.
2. The solubility of chromates falls down the group, so BaCrO4 has the lowest \( K_{sp} \).

Teacher's Note:
1. Smallest Ksp precipitates first.

Answer: (d) \( \left[\frac{10^{-4}}{1.6\times1.6}\right]^{1/3} \)

Explanation:
1. With solubility s and 80% dissociation, \( [Pb^{2+}]=0.8s \) and \( [Br^-]=1.6s \).
2. \( K_{sp}=0.8s(1.6s)^2=8\times10^{-5} \) gives \( s^3=\frac{8\times10^{-5}}{0.8\times1.6\times1.6}=\frac{10^{-4}}{1.6\times1.6} \).
3. So \( s=\left[\frac{10^{-4}}{1.6\times1.6}\right]^{1/3} \).

Teacher's Note:
1. Multiply each ion concentration by the degree of dissociation.

Answer: (b) \( \ce{CO2} \), \( \ce{H2CO3} \), \( \ce{HCO3-} \), \( \ce{CO3^{2-}} \)

Explanation:
1. Dissolved CO2 partly forms carbonic acid, which ionises in two steps.
2. The solution therefore contains CO2, H2CO3, HCO3- and CO3(2-) (plus H+).

Teacher's Note:
1. Diprotic acid: both anions are present, the second in tiny amount.

Answer: (b) \( 5\times10^{-9} \) M

Explanation:
1. After mixing in equal volumes each concentration becomes c/2.
2. No precipitation requires \( (c/2)^2\le6.25\times10^{-18} \), so \( c/2\le2.5\times10^{-9} \).
3. Hence \( c_{max}=5\times10^{-9} \) M.

Teacher's Note:
1. Account for the dilution on mixing before applying Ksp.

Answer: (d) 100 : 1

Explanation:
1. For equimolar solutions, \( \alpha=\frac{[H^+]}{C} \), so \( \alpha\propto[H^+] \).
2. \( [H^+] \) values are \( 10^{-3} \) and \( 10^{-5} \) M, giving a ratio of 100 : 1.

Teacher's Note:
1. At equal concentration, degree of dissociation is proportional to [H+].

Answer: (a) 6

Explanation:
1. Moles: KCN = 40 mmol, HCl = 20 mmol.
2. The reaction gives 20 mmol HCN and leaves 20 mmol \( \ce{CN-} \): a buffer with equal amounts.
3. \( pH=pK_a+\log1=6 \).

Teacher's Note:
1. Equal acid and conjugate base means pH = pKa.

Answer: (c) \( 5\times10^{-6} \) M

Explanation:
1. Total volume = 5 L, so \( [CH_3COO^-]=\frac{0.25}{5}=0.05 \) M (NaCl does not affect hydrolysis).
2. \( [OH^-]=\sqrt{K_bC}=\sqrt{5\times10^{-10}\times0.05}=5\times10^{-6} \) M.

Teacher's Note:
1. NaCl only dilutes the acetate.

Answer: (c) \( 5\times10^{-4} \) M

Explanation:
1. \( K_{sp}=4s^3=4\times(5\times10^{-6})^3=5\times10^{-16} \).
2. At pH 8, \( [OH^-]=10^{-6} \) M, so \( s'=\frac{K_{sp}}{[OH^-]^2}=\frac{5\times10^{-16}}{10^{-12}}=5\times10^{-4} \) M.

Teacher's Note:
1. Hold [OH-] fixed by the buffer and solve for [Pb2+].