JEE Chemistry Ionic Equilibrium MCQs Set 18

Answer: (a) NH3

Explanation:
1. The conjugate acid is formed by adding one proton to the base.
2. NH2- + H+ gives NH3.

Teacher's Note:
1. Conjugate acid = base + H+.
2. NH3 is also the conjugate base of NH4+.

Answer: (b) CH3COO-

Explanation:
1. A Bronsted acid must have a proton to donate.
2. CH3NH3+, H2O and HSO4- all can donate H+.
3. CH3COO- has no ionisable proton, it is only a base.

Teacher's Note:
1. Bronsted acid means proton donor.
2. Acetate is the conjugate base of acetic acid.

Answer: (c) III and IV

Explanation:
1. An amphiprotic species can both donate and accept a proton.
2. H2PO4- gives HPO4 2- or H3PO4, and HCO3- gives CO3 2- or H2CO3, so they are amphiprotic.
3. HPO3 2- has its hydrogen bonded to phosphorus and cannot donate it, and OH- is treated as only a base in water.

Teacher's Note:
1. Amphiprotic ions are the intermediate ions of polyprotic acids.
2. HPO3 2- is the ion of the dibasic phosphorous acid.

Answer: (b) H3O+

Explanation:
1. NH3 + H2O gives NH4+ + OH-, so [OH-] rises.
2. Since Kw = [H3O+][OH-] is constant, [H3O+] falls.

Teacher's Note:
1. Adding a base lowers [H3O+].
2. Kw stays constant at a fixed temperature.

Answer: (c) CH3+, -OC2H5

Explanation:
1. In (c) CH3+ (acid) combines with C2H5O- (base) to give CH3OC2H5, an ether, which has lone pairs on oxygen and is a Lewis base.
2. In (a) the adduct F3B-NH3 has no free lone pair on N and is not a base.
3. In (b) SiCl6 2- is a saturated complex whose chloride ligands are not basic in the sense used here.

Teacher's Note:
1. A Lewis base needs a free lone pair.
2. Adducts of BF3 are stable and are not bases.

Answer: (d) temperature is increased

Explanation:
1. Kw depends only on temperature.
2. The ionisation of water is endothermic, so a rise in temperature increases Kw.

Teacher's Note:
1. Adding acid or base does not change Kw.
2. Kw is \( 10^{-14} \) at 25 C and about \( 10^{-12} \) at 90 C.

Answer: (d) \( 10^{-13.4} \)

Explanation:
1. In pure water [H+] = [OH-] = \( 10^{-6.7} \).
2. Kw = [H+][OH-] = 10^-\( 6.7 \times 10^{-6} \).7 = \( 10^{-13.4} \).

Teacher's Note:
1. The pH of neutral water is below 7 above 25 C.
2. It is still neutral because [H+] = [OH-].

Answer: (c) \( 6.022 \times 10^{7} \)

Explanation:
1. 2NH3 gives NH4+ + NH2-, so [NH4+] = [NH2-] = square root of \( 10^{-30} \) = \( 10^{-15} \) mol/L.
2. In 100 cm3 the moles are \( 10^{-16} \).
3. Number of ions = \( 10^{-16} \) x \( 6.022 \times 10^{23} \) = \( 6.022 \times 10^{7} \).

Teacher's Note:
1. Self ionisation gives equal numbers of cations and anions.
2. Convert litres to the volume asked for.

Answer: (c) reduced by 1000 times

Explanation:
1. [H+] at pH 3 is \( 10^{-3} \) M and at pH 6 is \( 10^{-6} \) M.
2. The ratio is \( 10^{-3} \), so [H+] is reduced 1000 times.

Teacher's Note:
1. Each unit of pH changes [H+] by a factor of 10.
2. Higher pH means lower [H+].

Answer: (c) 9

Explanation:
1. 1/200 mol/m3 = 1/200 mmol/L = \( 5 \times 10^{-6} \) mol/L.
2. H2SO4 gives two H+ per molecule, so [H+] = \( 10^{-5} \) M and pH = 5.
3. pOH = 14 - 5 = 9.

Teacher's Note:
1. 1 mol/m3 equals \( 10^{-3} \) mol/L.
2. Use two H+ per molecule for sulphuric acid.

Answer: (b) 2

Explanation:
1. [H+] = C x alpha = 1.0 x 0.01 = \( 10^{-2} \) M.
2. pH = 2.

Teacher's Note:
1. For a monoprotic acid [H+] = C alpha.
2. Smaller ionisation gives a higher pH.

Answer: (c) \( 2 \times 10^{-3} \) M and \( 5 \times 10^{-12} \) M

Explanation:
1. [H+] = 0.1 x 0.02 = \( 2 \times 10^{-3} \) M.
2. [OH-] = Kw/[H+] = \( 10^{-14} \) / (\( 2 \times 10^{-3} \)) = \( 5 \times 10^{-12} \) M.

Teacher's Note:
1. Find [H+] first, then use Kw for [OH-].
2. The product [H+][OH-] stays \( 10^{-14} \).

Answer: (c) \( 3.85 \times 10^{-8} \)

Explanation:
1. Let the H+ from water be x, so [H+] = \( 2.23 \times 10^{-7} \) + x and [OH-] = x.
2. x(\( 2.23 \times 10^{-7} \) + x) = \( 10^{-14} \) gives x = \( 3.82 \times 10^{-8} \) mol/L, close to \( 3.85 \times 10^{-8} \).
3. For 1 L the moles are about \( 3.85 \times 10^{-8} \).

Teacher's Note:
1. Acid suppresses the ionisation of water.
2. Solve the quadratic when the acid is dilute.

Answer: (c) 1 M NaOH

Explanation:
1. 1 M NaOH is a strong base with [OH-] = 1 M and pH about 14.
2. 1 M NH3 is weak (pH about 11.6), distilled water is 7 and chlorine water is acidic.

Teacher's Note:
1. Strong bases fully ionise.
2. Chlorine water contains HCl and HOCl.

Answer: (b) 13

Explanation:
1. NaOH = 8/40 = 0.2 mol and H2SO4 = 4.9/98 = 0.05 mol, which gives 0.1 mol H+.
2. After neutralisation, excess OH- = 0.2 - 0.1 = 0.1 mol in 1 L.
3. pOH = 1, so pH = 13.

Teacher's Note:
1. Neutralise first and then find the excess.
2. H2SO4 supplies two H+ per mole.

Answer: (b) 2

Explanation:
1. Mixing portions of the same solution does not change its concentration, so it stays M/200 = 0.005 M.
2. [H+] = 2 x 0.005 = 0.01 M.
3. pH = 2.

Teacher's Note:
1. Mixing equal concentrations does not change the concentration.
2. H2SO4 gives two H+.

Answer: (d) 75 ml of M/5 HCl + 25 ml of M/5 NaOH

Explanation:
1. In (a) and (c) the base is in excess, so the solution is basic.
2. In (b) the excess acid is 1 mmol in 100 ml, so [H+] = 0.01 M and pH = 2.
3. In (d) HCl = 15 mmol and NaOH = 5 mmol, so excess H+ = 10 mmol in 100 ml, [H+] = 0.1 M and pH = 1.

Teacher's Note:
1. Find the excess acid or base after mixing.
2. Divide by the total volume.

Answer: (d) 10000

Explanation:
1. [H+] at pH 2 = \( 10^{-2} \) M and at pH 6 = \( 10^{-6} \) M.
2. Ratio = \( 10^{-2} \) / \( 10^{-6} \) = \( 10^{4} \) = 10000.

Teacher's Note:
1. A difference of 4 pH units means a factor of \( 10^{4} \).
2. The pH scale is logarithmic.

Answer: (c) \( 5.0 \times 10^{-15} \)

Explanation:
1. For H2A going to 2H+ + A2- the overall constant is the product of the two steps.
2. K = K1 x K2 = \( 1.0 \times 10^{-5} \) x \( 5.0 \times 10^{-10} \) = \( 5.0 \times 10^{-15} \).

Teacher's Note:
1. Overall constants of stepwise equilibria multiply.
2. For formation constants the same multiplication applies.

Answer: (a) 2

Explanation:
1. \( K_h = K_w/K_a = 10^{-14}/(2 \times 10^{-9}) = 5 \times 10^{-6} \).
2. C = 1/80 = 0.0125 M.
3. \( h = \sqrt{K_h/C} = \sqrt{5\times10^{-6}/0.0125} = \sqrt{4\times10^{-4}} = 0.02 \), which is 2%.

Teacher's Note:
1. Use h = square root of (Kh / C) for the salt of a weak acid.
2. Check that h is small.

Answer: (d) sodium acetate

Explanation:
1. Sodium acetate is the salt of a strong base and a weak acid, and acetate hydrolyses to give OH-.
2. Ammonium chloride and sulphate give acidic solutions, and ammonium acetate is nearly neutral.

Teacher's Note:
1. Strong base + weak acid gives a basic salt.
2. Weak base + strong acid gives an acidic salt.

Answer: (b) \( 1.8 \times 10^{9} \)

Explanation:
1. The second reaction is the first plus the reverse of water ionisation: CH3COOH + OH- = (CH3COOH + H2O) + (H+ + OH- going to H2O).
2. K = Ka / Kw = \( 1.8 \times 10^{-5} \) / \( 10^{-14} \) = \( 1.8 \times 10^{9} \).

Teacher's Note:
1. Add reactions and multiply their constants.
2. Neutralisation of a weak acid by a strong base goes essentially to completion.