Question

For hydrocyanic acid, HCN, Ka = 4.9 × 10-10. Calculate the pH of 0.20 M NaCN. What is the concentration of HCN in the solution? for the ph i got 10 is that right?

Answer 1

Answer : The concentration of HCN in the solution is $2.0\times 10^{-3}M$

Explanation :

First we have to calculate the value of $pK_a$.

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (4.9\times 10^{-10})$

$pK_a=10-\log (4.9)$

$pK_a=9.3$

Now we have to calculate the pH of the solution.

The hydrolysis reaction will be,

$NaCN+H_2O\rightarrow HCN+NaOH$

Formula used :

$pH=7+\frac{1}{2}[pKa+\log C]$

$pH=7+\frac{1}{2}[9.3+\log (0.20)]$

$pH=11.3$

Now we have to calculate the concentration of HCN in the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[NaCN]}{[HCN]}$

Now put all the given values in this expression, we get:

$11.3=9.3+\log (\frac{0.20}{[HCN]})$

$[HCN]=2.0\times 10^{-3}M$

Therefore, the concentration of HCN in the solution is $2.0\times 10^{-3}M$

Answer 2

By using the ICE table :

initial    0.2 M            0           0
change -X                 + X           +X
Equ     (0.2 -X)            X               X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X 
∴X ≈ 10^-6
∴[HCN] = 10^-6

and PH = -㏒[H+]
             = -㏒ 10^-6
             = 6