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

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Step-by-step 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 = ([H_(3)O^(+)][CN^(-)])/([HCN]) = 6.17 \cdot 10^(-10)`

`6.17 \cdot 10^(-10) = (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:

`\% = (x)/([HCN]) * 100`

`\% = (1.75 \cdot 10^(-5) M)/(0.5 M) * 100`

`\% = 3.5 \cdot 10^(-3) \%`

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

I hope it helps you!