Question

Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 10-5, pH = 4.75, % rxn = 3.5 x 10-3%)

Answer 1

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!