JEE Chemistry Ionic Equilibrium MCQs Set 20

Answer: (d) NH4OH

Explanation:
1. NaOH, RbOH and KOH are strong bases and give a sharp end point with phenolphthalein.
2. NH4OH is a weak base and the equivalence point with HCl is acidic, so phenolphthalein cannot show it.

Teacher's Note:
1. A weak base with a strong acid needs methyl orange.
2. Phenolphthalein suits only basic end points.

Answer: (c) oxalic acid and KMnO4

Explanation:
1. Phenolphthalein is an acid-base indicator and works in neutralisation reactions.
2. Oxalic acid with KMnO4 is a redox titration, where KMnO4 acts as its own indicator.

Teacher's Note:
1. Redox titrations use different indicators.
2. KMnO4 is self-indicating.

Answer: (d) 108y5

Explanation:
1. A2X3 gives 2A3+ + 3X2-, so [A3+] = 2y and [X2-] = 3y.
2. Ksp = (2y)² (3y)³ = 4 x 27 \( y^{5} \) = 108 \( y^{5} \).

Teacher's Note:
1. Ksp = (a s)^a (b s)^b for AaBb.
2. For A2B3 it is 108 \( s^{5} \).

Answer: (d) 8

Explanation:
1. HgSO4 is a 1:1 salt, so s = square root of Ksp = square root of \( 6.4 \times 10^{-5} \) = \( 8 \times 10^{-3} \) mol/L.
2. 1 m3 = 1000 L, so s = 8 mol/m3.

Teacher's Note:
1. Convert mol/L to mol/m3 by multiplying by 1000.
2. For a 1:1 salt s = square root of Ksp.

Answer: (c) \( 1 \times 10^{-10} \)

Explanation:
1. Solubility = \( 1.43 \times 10^{-3} \) g/L = \( 1.43 \times 10^{-3} \)/143 = \( 10^{-5} \) mol/L.
2. Ksp = \( s^{2} \) = \( 10^{-10} \).

Teacher's Note:
1. Convert g/100 ml to mol/L first.
2. For AgCl Ksp = s2.

Answer: (c) (1/54)(\( M^{4} \)/\( x^{4} \))

Explanation:
1. Molar solubility s = x/M. A3B2 gives 3A2+ + 2B3-, so [B3-] = 2s = 2x/M.
2. Ksp = (3s)³ (2s)² = 108 \( s^{5} \) = 108 \( x^{5} \)/\( M^{5} \).
3. Ratio = (2x/M) / (108 \( x^{5} \)/\( M^{5} \)) = \( M^{4} \) / (54 \( x^{4} \)).

Teacher's Note:
1. Write both quantities in terms of s.
2. Ksp = 108 \( s^{5} \) for A3B2.

Answer: (b) 100

Explanation:
1. In pure water s = (Ksp/4)^(1/3) = (\( 8 \times 10^{-12} \))^(1/3) = \( 2 \times 10^{-4} \) M.
2. In KF, [F-] = \( 4 \times 10^{-3} \) M and s = Ksp/[F-]² = \( 3.2 \times 10^{-11} \) / \( 1.6 \times 10^{-5} \) = \( 2 \times 10^{-6} \) M.
3. The decrease = \( 2 \times 10^{-4} \) / \( 2 \times 10^{-6} \) = 100 times.

Teacher's Note:
1. A common ion reduces solubility.
2. Use Ksp = s [F-]² in the salt solution.

Answer: (d) 0.05 M NH3

Explanation:
1. Na2CO3 and AgNO3 add a common ion (CO3 2- or Ag+) and reduce the solubility.
2. Pure water gives the normal solubility.
3. NH3 forms the complex [Ag(NH3)2]+ with Ag+, which removes Ag+ from the solution and increases the solubility, so it is maximum in NH3.

Teacher's Note:
1. Complex formation increases solubility.
2. Common ions decrease it.

Answer: (b) a precipitate will not form

Explanation:
1. At pH 9, [OH-] = \( 10^{-5} \) M.
2. Ionic product = [Mg2+][OH-]² = 10^-\( 3 \times 10^{-10} \) = \( 10^{-13} \).
3. This is less than Ksp (\( 8.9 \times 10^{-12} \)), so no precipitate forms.

Teacher's Note:
1. Precipitation occurs only if the ionic product exceeds Ksp.
2. Convert pH to [OH-] first.

Answer: (b) 0.328 g

Explanation:
1. Maximum [PO4 3-] without precipitation: [PO4 3-]² = Ksp / [Ba2+]³ = \( 4 \times 10^{-23} \) / \( 10^{-15} \) = \( 4 \times 10^{-8} \), so [PO4 3-] = \( 2 \times 10^{-4} \) M.
2. Moles in 10 L = \( 2 \times 10^{-3} \).
3. Mass = \( 2 \times 10^{-3} \) x 164 = 0.328 g.

Teacher's Note:
1. Use the highest concentration that keeps the ionic product at Ksp.
2. The molar mass of Na3PO4 is 164 g/mol.

Answer: (c) \( 4 \times 10^{-15} \)

Explanation:
1. AB2 gives A2+ + 2B-, so [A2+] = s and [B-] = 2s.
2. Ksp = s(2s)² = 4s³ = 4 x (\( 10^{-5} \))³ = \( 4 \times 10^{-15} \).

Teacher's Note:
1. Ksp = 4s³ for any AB2 or A2B salt.
2. Write the ion concentrations in terms of s.

Answer: (c) 4x³

Explanation:
1. Mg(OH)2 gives Mg2+ + 2OH-, so [Mg2+] = x and [OH-] = 2x.
2. Ksp = x (2x)² = 4x³.

Teacher's Note:
1. The hydroxide concentration is twice the solubility.
2. Ksp = 4s³ for Mg(OH)2.

Answer: (d) s = (Ksp/256)^(1/5)

Explanation:
1. MX4 gives M4+ + 4X-, so [M4+] = s and [X-] = 4s.
2. Ksp = s (4s)⁴ = 256 \( s^{5} \).
3. So s = (Ksp/256)^(1/5).

Teacher's Note:
1. Raise the stoichiometric number to its own power.
2. The exponent of s is the total number of ions.

Answer: (a) \( 1.0 \times 10^{-4} \) M

Explanation:
1. For MX2, Ksp = 4s³ = \( 4 \times 10^{-12} \), so \( s^{3} \) = \( 10^{-12} \).
2. s = \( 10^{-4} \) M, and [M2+] = s = \( 1.0 \times 10^{-4} \) M.

Teacher's Note:
1. [M2+] equals s while [X-] equals 2s.
2. Take the cube root carefully.

Answer: (b) \( 2.83 \times 10^{-3} \) g

Explanation:
1. s = square root of Ksp = \( 10^{-4} \) mol/L.
2. In 100 ml the moles = \( 10^{-5} \).
3. Mass = \( 10^{-5} \) x 283 = \( 2.83 \times 10^{-3} \) g.

Teacher's Note:
1. Scale the moles to the volume given.
2. Multiply by the molar mass at the end.

Answer: (d) \( 5.1 \times 10^{-5} \) M

Explanation:
1. Precipitation starts when [Ba2+][CO3 2-] = Ksp.
2. [Ba2+] = \( 5.1 \times 10^{-9} \) / \( 1.0 \times 10^{-4} \) = \( 5.1 \times 10^{-5} \) M.

Teacher's Note:
1. The ionic product just reaches Ksp at the onset.
2. Divide Ksp by the known ion.

Answer: (c) \( 1.2 \times 10^{-9} \) g

Explanation:
1. [Br-] needed = Ksp/[Ag+] = \( 5.0 \times 10^{-13} \) / 0.05 = \( 10^{-11} \) M.
2. For 1 L the moles = \( 10^{-11} \), so mass = \( 10^{-11} \) x 120 = \( 1.2 \times 10^{-9} \) g.

Teacher's Note:
1. Precipitation begins when the ionic product equals Ksp.
2. Convert moles to grams with the molar mass.

Answer: (c) \( \sqrt[4]{1.6 \times 10^{-30}/27} \)

Explanation:
1. Cr(OH)3 gives Cr3+ + 3OH-, so [Cr3+] = s and [OH-] = 3s.
2. Ksp = s (3s)³ = 27 \( s^{4} \), so s = (Ksp/27)^(1/4).

Teacher's Note:
1. Ksp = 27 \( s^{4} \) for A B3 salts.
2. Take the fourth root.

Answer: (a) \( 1 \times 10^{-10} \)

Explanation:
1. [H+] = [A-] = \( 10^{-5} \) M and [HA] is about 1.0 M.
2. Ka = (\( 10^{-5} \))²/1.0 = \( 10^{-10} \).

Teacher's Note:
1. For a weak acid [H+] = square root of (Ka x C).
2. Ka = [H+]² / C.

Answer: (d) \( 1 \times 10^{-2} \) M Ca2+ and \( 1 \times 10^{-3} \) M F-

Explanation:
1. A precipitate forms when [Ca2+][F-]² is greater than \( 1.7 \times 10^{-10} \).
2. (a) 10^-\( 2 \times 10^{-10} \) = \( 10^{-12} \), (b) 10^-\( 4 \times 10^{-8} \) = \( 10^{-12} \), (c) 10^-\( 3 \times 10^{-10} \) = \( 10^{-13} \), all below Ksp.
3. (d) 10^-\( 2 \times 10^{-6} \) = \( 10^{-8} \), which is greater than Ksp, so CaF2 precipitates.

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

Answer: (d) \( 1 \times 10^{-5} \)

Explanation:
1. [H+] = \( 10^{-3} \) M.
2. Ka = [H+]² / C = \( 10^{-6} \) / 0.1 = \( 10^{-5} \).

Teacher's Note:
1. Use Ka = [H+]2/C for a weak acid.
2. The dissociation is small compared with 0.1 M.

Answer: (d) 9.0 L

Explanation:
1. pH 1 means 0.1 M and pH 2 means 0.01 M, so the solution has to be diluted 10 times.
2. The final volume is 10 L, so 9 L of water are to be added.

Teacher's Note:
1. Use M1V1 = M2V2.
2. Water added = final volume minus initial volume.