Answer:
The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44
Explanation:
Given: The base dissociation constant: [tex]K_{b}[/tex] = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M
Also, water dissociation constant: [tex]K_{w}[/tex] = 1 × 10⁻¹⁴
The acid dissociation constant ([tex]K_{a}[/tex]) for the weak acid (BH⁺) can be calculated by the equation:
[tex]K_{a}. K_{b} = K_{w}[/tex]
[tex]\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}[/tex]
[tex]\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}[/tex]
Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:
Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+
Initial: 0.1 M x x
Change: -x +x +x
Equilibrium: 0.1 - x x x
The acid dissociation constant: [tex]K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}[/tex]
[tex]\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}[/tex]
[tex]\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}[/tex]
[tex]As, x << 0.1[/tex]
[tex]\Rightarrow 0.1 - x = 0.1[/tex]
[tex]\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }[/tex]
[tex]\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}[/tex]
[tex]\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}[/tex]
Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M
Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44
Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44