Answer:
0.002 %
Explanation:
Given that:
[tex]K_{a}=4.0\times 10^{-10}[/tex]
Concentration = 1.0 M
Consider the ICE take for the dissociation of Hydrocyanic acid as:
HCN ⇄ H⁺ + CN⁻
At t=0 1.0 - -
At t =equilibrium (1.0-x) x x
The expression for dissociation constant of Hydrocyanic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CN}^- \right ]}{[HCN]}[/tex]
[tex]4.0\times 10^{-10}=\frac {x^2}{1.0-x}[/tex]
x is very small, so (1.0 - x) ≅ 1.0
Solving for x, we get:
x = 2×10⁻⁵ M
Percentage ionization = [tex]\frac {2\times 10^{-5}}{1.0}\times 100=0.002 \%[/tex]
