Question

Hydrocyanic acid has a Ka of 4.0

x10-10. What is the percent ofionization of
a 1.0 molar solution?

Answer 1

0.002 %

Step-by-step explanation:

Given that:

`K_(a)=4.0* 10^(-10)`

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:

`K_(a)=\frac {\left [ H^(+) \right ]\left [ {CN}^- \right ]}{[HCN]}`

`4.0* 10^(-10)=\frac {x^2}{1.0-x}`

x is very small, so (1.0 - x) ≅ 1.0

Solving for x, we get:

x = 2×10⁻⁵ M

Percentage ionization =
`\frac {2* 10^(-5)}{1.0}* 100=0.002 \%`