Question

What is (h+) of a solution that is 0.1 m hcn and 0.2m nacn (ka for hcn =6.2×10'10)?

Answer 1

Answer : The  concentration of hydrogen ion is $3.2\times 10^{-10}M$

Explanation : Given,

$K_a=6.2\times 10^{-10}$

Concentration of HCN = 0.1 M

Concentration of NaCN = 0.2 M

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:

$pK_a=-\log (6.2\times 10^{-10})$

$pK_a=10-\log (6.2)$

$pK_a=9.2$

Now we have to calculate the pH of buffer.

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:

$pH=9.2+\log (\frac{0.2}{0.1})$

$pH=9.5$

The pH of buffer is 9.5.

Now we have tom calculate the concentration of hydrogen ion.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$9.5=-\log [H^+]$

$[H^+]=3.2\times 10^{-10}M$

Therefore, the concentration of hydrogen ion is $3.2\times 10^{-10}M$