JEE Chemistry Equilibrium MCQs Set 04

Answer: (c) \(1.6\times10^{-9}\ \text{M}\)

Explanation:
1. Common ion effect: \([Cl^-]=0.1\) M from HCl (the chloride from AgCl is negligible)
2. \(K_{sp}=s\times0.1\)
3. \(s=\frac{1.6\times10^{-10}}{0.1}=1.6\times10^{-9}\) M

Teacher's Note:
1. The solubility falls in a solution with a common ion. Compare with the pure-water value \(\sqrt{K_{sp}}=1.26\times10^{-5}\) M.

Answer: (c) \(27S^4\)

Explanation:
1. \(\ce{Na3PO4(s) <=> 3Na^+ + PO4^3-}\)
2. \([Na^+]=3S\) and \([PO_4^{3-}]=S\)
3. \(K_{sp}=(3S)^3\cdot S=27S^4\)

Teacher's Note:
1. Write the dissociation, raise each ion concentration to its stoichiometric coefficient and substitute in terms of S. The general form for \(A_xB_y\) is \(x^xy^yS^{x+y}\).

Answer: (b) \(5.0\times10^{-7}\)

Explanation:
1. \(pH=12\) gives \(pOH=2\) and \([OH^-]=10^{-2}\) M
2. \([Ba^{2+}]=\tfrac12[OH^-]=5\times10^{-3}\) M
3. \(K_{sp}=(5\times10^{-3})(10^{-2})^2=5\times10^{-7}\)

Teacher's Note:
1. Barium hydroxide gives two hydroxide ions per formula unit, so the barium concentration is half the hydroxide concentration.

Answer: (b) \(6912x^7\)

Explanation:
1. \(\ce{Zr3(PO4)4 <=> 3Zr^4+ + 4PO4^3-}\)
2. \([Zr^{4+}]=3x\) and \([PO_4^{3-}]=4x\)
3. \(K_{sp}=(3x)^3(4x)^4=27\times256\,x^7=6912x^7\)

Teacher's Note:
1. Raise each concentration to the power of its coefficient: \(3^3\times4^4=27\times256=6912\).

Answer: (a) \(\ce{ZnS}\) (\(K_{sp}=2\times10^{-10}\))

Explanation:
1. For salts of the same type (\(\ce{MS}\)) solubility is \(\sqrt{K_{sp}}\), so a larger \(K_{sp}\) means higher solubility
2. The largest value is \(2\times10^{-10}\) for ZnS

Teacher's Note:
1. Compare solubility products directly only for salts that give the same number of ions. Here all four give one cation and one anion.

Answer: (d) \(\ce{HCOOH}\) + NaOH (with the acid in excess)

Explanation:
1. An acidic buffer is a weak acid with its salt
2. Part-neutralising excess \(\ce{HCOOH}\) with NaOH leaves \(\ce{HCOOH}\) and \(\ce{HCOONa}\)
3. Option 1 would be a basic buffer, and options 2 and 3 do not contain a weak acid with its salt

Teacher's Note:
1. An acidic buffer has pH below 7 and resists change when small amounts of acid or alkali are added. It is made from a weak acid and its conjugate base.

Answer: (c) 4.0

Explanation:
1. \(pH=pK_a+\log\frac{[\text{salt}]}{[\text{acid}]}\)
2. Buffer capacity is maximum when [salt] = [acid]
3. Then \(pH=pK_a=4\).

Teacher's Note:
1. A buffer works best when the pH is close to the pKa of the weak acid, because both components are then present in equal amounts.

Answer: (d) 500 mL of 0.2 N \(\ce{CH3COOH}\) + 500 mL of 0.1 N NaOH

Explanation:
1. Count milliequivalents: 0.1 N x 500 mL = 50 meq; 0.2 N x 500 mL = 100 meq
2. Option 4: 100 meq acid + 50 meq NaOH leaves 50 meq acid and forms 50 meq salt, so it is a buffer
3. Option 1 gives only the salt, option 3 leaves excess NaOH and option 2 has a strong acid

Teacher's Note:
1. Mix a weak acid with a smaller amount of strong base and you get a buffer. Complete neutralisation gives only the salt and no buffer.

Answer: (d) \(10^{-6}\)

Explanation:
1. \(K_h=\frac{K_w}{K_a}\)
2. \(K_a=\frac{10^{-14}}{10^{-8}}=10^{-6}\)

Teacher's Note:
1. The hydrolysis constant of a salt of a weak acid is the ratio of Kw to Ka. Hydrolysis constant does not depend on the concentration.

Answer: (b) 11

Explanation:
1. \(\ce{CN^- + H2O <=> HCN + OH^-}\) with \(K_b=10^{-5}\)
2. \([OH^-]=\sqrt{K_bC}=\sqrt{10^{-5}\times0.1}=10^{-3}\) M
3. \(pOH=3\), so \(pH=11\)

Teacher's Note:
1. A salt of a strong base and a weak acid gives a basic solution. Use \(pH=14-pOH\) after finding \([OH^-]\).