JEE Chemistry Ionic Equilibrium MCQs Set 24

Answer: (d) 5.0

Explanation:
1. 50 mL of acid is needed and the total volume is 100 mL, so NH4Cl is 0.2 M.
2. Kh = \( 10^{-14} \)/\( 2 \times 10^{-5} \) = \( 5 \times 10^{-10} \) and [H+] = square root of (\( 5 \times 10^{-10} \) x 0.2) = \( 10^{-5} \) M, so pH = 5.

Teacher's Note:
1. The end point is acidic for a weak base with a strong acid.
2. Use salt hydrolysis.

Answer: (a) 8.70

Explanation:
1. The salt is 5 mmol in 100 ml, so 0.05 M. Kh = \( 5 \times 10^{-10} \).
2. [OH-] = square root of (\( 5 \times 10^{-10} \) x 0.05) = \( 5 \times 10^{-6} \) M, so pOH = 5.30 and pH = 8.70.

Teacher's Note:
1. The equivalence point of a weak acid with a strong base is above 7.
2. Use salt hydrolysis.

Answer: (d) 80 %

Explanation:
1. Half ionised at pH 3 means pKIn = 3, so Ka = \( 10^{-3} \).
2. [HIn]/[In-] = [H+]/Ka = \( 4 \times 10^{-3} \)/\( 10^{-3} \) = 4, so unionised = 4/5 = 80 %.

Teacher's Note:
1. At the half way point pH = pKIn.
2. Ratio of 4 : 1 means 80 % : 20 %.

Answer: (b) 85.71 %

Explanation:
1. [In-]/[HIn] = Ka/[H+] = \( 6 \times 10^{-5} \)/\( 10^{-5} \) = 6.
2. Basic form = 6/7 = 85.71 %.

Teacher's Note:
1. The ratio of ionised to unionised form is Ka/[H+].
2. Convert the ratio to a percentage of the total.

Answer: (d) 0.954

Explanation:
1. At 75% red the ratio base/acid = 1/3 and at 75% blue it is 3.
2. The pH changes by log 3 - log(1/3) = 2 x 0.4771 = 0.954.

Teacher's Note:
1. pH = pKa + log (base/acid).
2. The change is twice log 3.

Answer: (a) QX2

Explanation:
1. MX: s = square root of Ksp = \( 2 \times 10^{-9} \) M.
2. QX2: s = (Ksp/4)^(1/3) = (\( 10^{-18} \))^(1/3) = \( 10^{-6} \) M.
3. QX2 has the higher solubility.

Teacher's Note:
1. Compare solubilities, not Ksp values, for salts of different types.
2. An AB2 salt has Ksp = 4s³.

Answer: (a) \( 4 \times 10^{-8} \)

Explanation:
1. s = 0.0608/304 = \( 2 \times 10^{-4} \) mol/L.
2. Ksp = \( s^{2} \) = \( 4 \times 10^{-8} \).

Teacher's Note:
1. Convert g/L to mol/L first.
2. A 1:1 salt has Ksp = \( s^{2} \).

Answer: (b) \( 3.2 \times 10^{-11} \)

Explanation:
1. [Ca2+] = \( 2 \times 10^{-4} \) M.
2. Ksp = [Ca2+][F-]² = \( 2 \times 10^{-4} \) x (\( 4 \times 10^{-4} \))² = \( 3.2 \times 10^{-11} \).

Teacher's Note:
1. [Ca2+] is half of [F-].
2. Square the fluoride concentration.

Answer: (b) \( 2.56 \times 10^{-16} \)

Explanation:
1. s = \( 2.4 \times 10^{-4} \) g/L / 60 = \( 4 \times 10^{-6} \) mol/L.
2. Ksp = 4 \( s^{3} \) = 4 x (\( 4 \times 10^{-6} \))³ = \( 2.56 \times 10^{-16} \).

Teacher's Note:
1. Convert to g/L first, then to mol/L.
2. An ML2 salt has Ksp = 4s³.

Answer: (d) \( 1.6 \times 10^{-6} \) mol/L

Explanation:
1. Let [Ag+] = x. [SCN-] = \( 3.2 \times 10^{-12} \)/x and [Br-] = \( 8 \times 10^{-13} \)/x.
2. Charge balance: x = [SCN-] + [Br-] = \( 4 \times 10^{-12} \)/x, so x = \( 2 \times 10^{-6} \) M.
3. [SCN-] = \( 3.2 \times 10^{-12} \)/\( 2 \times 10^{-6} \) = \( 1.6 \times 10^{-6} \) mol/L.

Teacher's Note:
1. The same Ag+ is shared by both salts.
2. Use the charge balance.

Answer: (a) \( 3 \times 10^{-3} \) M

Explanation:
1. [Mg2+] = \( 9.5 \times 10^{-9} \)/[F-]² and [Sr2+] = \( 4 \times 10^{-9} \)/[F-]².
2. Charge balance: [F-] = 2([Mg2+] + [Sr2+]) = 2 x \( 13.5 \times 10^{-9} \)/[F-]².
3. [F-]³ = \( 2.7 \times 10^{-8} \), so [F-] = \( 3 \times 10^{-3} \) M.

Teacher's Note:
1. Both salts share fluoride.
2. Use [F-] = 2([Mg2+] + [Sr2+]).

Answer: (a) \( 2 \times 10^{-9} \) M

Explanation:
1. [CN-] = 0.4 M, so s = Ksp/[CN-] = \( 8 \times 10^{-10} \)/0.4 = \( 2 \times 10^{-9} \) M.

Teacher's Note:
1. Neglecting the complex, only the common ion effect applies.
2. In reality complex formation raises the solubility.

Answer: (d) \( 4 \times 10^{-2} \) M

Explanation:
1. CN- is protonated: \( S^{2} \) = Ksp(1 + [H+]/Ka) = \( 8 \times 10^{-10} \) x (1 + \( 10^{-3} \)/\( 5 \times 10^{-10} \)).
2. = \( 8 \times 10^{-10} \) x \( 2 \times 10^{6} \) = \( 1.6 \times 10^{-3} \), so S = \( 4 \times 10^{-2} \) M.

Teacher's Note:
1. A weak acid anion dissolves more at low pH.
2. Use \( S^{2} \) = Ksp (1 + [H+]/Ka).

Answer: (b) Only in (b)

Explanation:
1. (a) Ionic product = \( 10^{-3} \) x (\( 10^{-5} \))² = \( 10^{-13} \), less than \( 1.2 \times 10^{-11} \), so no precipitate.
2. (b) Ionic product = \( 10^{-3} \) x (\( 10^{-3} \))² = \( 10^{-9} \), greater than Ksp, so a precipitate forms.

Teacher's Note:
1. Compare the ionic product with Ksp.
2. Square the hydroxide concentration.

Answer: (d) 0.284 g

Explanation:
1. [SO4 2-] needed = \( 8 \times 10^{-8} \)/\( 2 \times 10^{-5} \) = \( 4 \times 10^{-3} \) M.
2. Moles in 0.5 L = \( 2 \times 10^{-3} \), so mass = \( 2 \times 10^{-3} \) x 142 = 0.284 g.

Teacher's Note:
1. Find the sulphate needed at the onset of precipitation.
2. The molar mass of Na2SO4 is 142 g/mol.

Answer: (b) 0.209 M

Explanation:
1. After the first step [H+] = [HSO4-] = 0.20 M.
2. For the second step: x(0.20 + x)/(0.20 - x) = \( 10^{-2} \), giving \( x^{2} \) + 0.21x - 0.002 = 0 and x = \( 9.13 \times 10^{-3} \).
3. [H+] = 0.20 + 0.0091 = 0.209 M.

Teacher's Note:
1. The second ionisation is partial.
2. Solve the quadratic equation.

Answer: (b) 10.52

Explanation:
1. Kh = \( 10^{-14} \)/\( 0.6 \times 10^{-10} \) = \( 1.67 \times 10^{-4} \).
2. \( x^{2} \)/(\( 10^{-3} \) - x) = \( 1.67 \times 10^{-4} \) gives x = [OH-] = \( 3.33 \times 10^{-4} \) M, so pOH = 3.48 and pH = 10.52.

Teacher's Note:
1. Kh is large compared with C, so the simple square root formula is not accurate.
2. Solve the quadratic.

Answer: (d) \( 3.73 \times 10^{-2} \) M

Explanation:
1. PO4 3- + H2O gives HPO4 2- + OH-, with Kb = \( 10^{-14} \)/\( 4.5 \times 10^{-13} \) = \( 2.22 \times 10^{-2} \).
2. \( x^{2} \)/(0.1 - x) = \( 2.22 \times 10^{-2} \) gives \( x^{2} \) + 0.0222x - 0.00222 = 0 and x = \( 3.73 \times 10^{-2} \) M.

Teacher's Note:
1. Phosphate is a fairly strong base.
2. The approximation x much smaller than 0.1 fails.

Answer: (a) 9.60

Explanation:
1. HPO4 2- is amphiprotic, so pH = (pK2 + pK3)/2.
2. pK2 = 7.21 and pK3 = 12.0, so pH = (7.21 + 12.0)/2 = 9.60.

Teacher's Note:
1. Use the average of the two relevant pK values.
2. The pH does not depend on the concentration.

Answer: (b) \( 1 \times 10^{-3} \) M

Explanation:
1. HCN = 75 mmol needs 50 ml KOH, so the total volume is 200 ml and [CN-] = 0.375 M.
2. Kh = \( 10^{-14} \)/\( 3.75 \times 10^{-9} \) = \( 2.67 \times 10^{-6} \).
3. [HCN] = [OH-] = square root of (\( 2.67 \times 10^{-6} \) x 0.375) = \( 10^{-3} \) M.

Teacher's Note:
1. The hydrolysis of the salt re-forms a little HCN.
2. [HCN] equals [OH-].

Answer: (c) \( 2.8 \times 10^{-3} \) mol

Explanation:
1. AgBr + 2NH3 gives Ag(NH3)2+ + Br-, with K = Ksp x Kf = \( 5 \times 10^{-5} \).
2. \( s^{2} \)/(0.4 - 2s)² = \( 5 \times 10^{-5} \), so s/(0.4 - 2s) = \( 7.07 \times 10^{-3} \).
3. s = 0.4 x \( 7.07 \times 10^{-3} \)/(1 + 0.0141) = \( 2.8 \times 10^{-3} \) mol.

Teacher's Note:
1. Complex formation increases the solubility.
2. Combine Ksp and Kf into one equilibrium constant.