Answer
$pH = 5.694$
Work Step by Step
- Since $Cl^-$ doesn't affect the water equilibrium, we just need to calculate the pH for the $CH_3N{H_3}^+$.
1. Since $CH_3N{H_3}^+$ is the conjugate acid of $CH_3NH_2$ , we can calculate its kb by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 4.4\times 10^{- 4} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 4.4\times 10^{- 4}}$
$K_a = 2.273\times 10^{- 11}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [CH_3NH_2] = x$
-$[CH_3N{H_3}^+] = [CH_3N{H_3}^+]_{initial} - x = 0.18 - x$
For approximation, we consider: $[CH_3N{H_3}^+] = 0.18M$
3. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3NH_2]}{ [CH_3N{H_3}^+]}$
$Ka = 2.273 \times 10^{- 11}= \frac{x * x}{ 0.18}$
$Ka = 2.273 \times 10^{- 11}= \frac{x^2}{ 0.18}$
$ 4.091 \times 10^{- 12} = x^2$
$x = 2.023 \times 10^{- 6}$
Percent dissociation: $\frac{ 2.023 \times 10^{- 6}}{ 0.18} \times 100\% = 0.001124\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [CH_3NH_2] = x = 2.023 \times 10^{- 6}M $
And, since 'x' has a very small value (compared to the initial concentration): $[CH_3N{H_3}^+] \approx 0.18M$
4. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.023 \times 10^{- 6})$
$pH = 5.694$
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