JEE Chemistry Ionic Equilibrium Notes Set 03

Hydrolysis of Salts

Highly charged small cations such as Fe3+ and Al3+ hydrolyse strongly, and carbonate, sulphide and cyanide ions hydrolyse to give basic solutions. An amphiprotic anion such as HCO3- or H2PO4- gives \( pH = (pK_1 + pK_2)/2 \), independent of concentration.

Buffer Solutions

1. A buffer resists a change in pH on adding small amounts of acid or base, or on dilution. It contains a weak acid with its salt (acidic buffer) or a weak base with its salt (basic buffer).
2. Acidic buffer: \( pH = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} \). Basic buffer: \( pOH = pK_b + \log\dfrac{[\text{salt}]}{[\text{base}]} \) (Henderson's equation).
3. Buffer capacity is the moles of acid or base needed per litre to change the pH by one unit. It is highest when [salt] = [acid], where \( pH = pK_a \). A useful range is \( pK_a \pm 1 \).
4. A buffer can be formed in situ by partly neutralising a weak acid with a strong base, or a weak base with a strong acid. Work out the moles after reaction and then use the Henderson equation.
5. Mixtures such as strong acid with its own salt (HCl and NaCl) or a weak acid with a strong base in excess are not buffers.

Indicators and Acid-Base Titrations

An acid indicator HIn is a weak acid whose ionised and unionised forms have different colours. \( pH = pK_{In} + \log\dfrac{[In^-]}{[HIn]} \). The colour change takes place over about \( pK_{In} \pm 1 \).

During titration of a weak acid with a strong base, the pH at half the equivalence volume equals pKa. After the equivalence point the pH is controlled by the excess strong base.

Solubility and Solubility Product

1. For a sparingly soluble salt \( A_xB_y \rightleftharpoons xA^{y+} + yB^{x-} \), \( K_{sp} = (xs)^x(ys)^y = x^xy^y s^{x+y} \), where s is the molar solubility.
2. So \( K_{sp} = s^2 \) for 1:1 salts, \( 4s^3 \) for AB2 or A2B, \( 27s^4 \) for AB3 or A3B, and \( 108s^5 \) for A2B3 or A3B2.
3. The ionic product Q compares with Ksp: precipitation occurs if \( Q > K_{sp} \), the solution is saturated if \( Q = K_{sp} \), and the solution is unsaturated if \( Q < K_{sp} \).
4. Common ion effect: a common ion lowers the solubility. For example the solubility of AgCl in c M NaCl is Ksp / c.
5. Solubility is raised by complex formation (AgCl dissolves in NH3) and, for salts of weak acids (sulphides, carbonates, hydroxides, cyanides), by lowering the pH.
6. Selective precipitation: the ion whose salt needs the smaller concentration of the precipitating agent precipitates first. Simultaneous solubility of two salts with a common ion is found from the charge balance.
7. Solubility cannot be compared by Ksp alone unless the salts have the same formula type.

Worked Examples

1. Solubility of CaF2, \( K_{sp} = 3.2\times10^{-11} \): \( s = (K_{sp}/4)^{1/3} = 2\times10^{-4} \) M.
2. Mg(OH)2 (\( K_{sp} = 10^{-11} \)) starts precipitating from 0.001 M Mg2+ when \( [OH^-] = \sqrt{10^{-11}/10^{-3}} = 10^{-4} \) M, that is, at pH 10.
3. A buffer of 0.1 M CH3COOH and 0.1 M CH3COONa has pH = 4.74. Adding 3.33 millimoles of NaOH to 100 mL of it raises the pH by 0.3.