BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, which is neither an acid nor a base. Find the pH of 0.100 M BH+ClO4-.

Respuesta :

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

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