Question

Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10 DONE

Answer 1

Answer:

The pH of the solution is 5.31.

Explanation:

Let "$\alpha$ is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  $HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}$

Initial                  C                         0            0

Equilibrium        c(1- $\alpha$)              c$\alpha$ c$\alpha$

Dissociation constant = $Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}$

$=\frac{c\alpha^{2}}{(1-\alpha)}$

In this case weak acids $\alpha$ is very small so, (1-$\alpha$ ) is taken as 1.

$Ka=C\alpha^{2}$

$\alpha=\sqrt\frac{ka}{c}$

From the given the concentration = 0.050 M

Substitute the given value.

$\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}$

$[H_{3}O^{+}]=c\alpha$

$[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}$

$pH= -log[H_{3}O^{+}]$

$=-log[4.9\times10^{-6}]$

$=6-log 4.9= 5.31$

Therefore, The pH of the solution is 5.31.

Answer 2

Answer:

5.3

The next one is 11

Explanation: