The pH of a 0.02M $NH_4Cl$ solution will be [given $K_b(NH_4OH) = 10^{-5}$ and log2=0.301]

Question

What is the pH of a 0.1 M NaOH solution?

Answer 1

To determine the pH of a 0.02 M $NH_4Cl$ solution, we need to understand that $NH_4Cl$ is a salt of a weak base ($NH_4OH$) and a strong acid ($HCl$). The solution will be acidic, and we need to find the concentration of hydrogen ions $[\text{H}^+]$ to calculate the pH.

We know that:

$K_b$ for $NH_4OH = 10^{-5}$
To find the $K_a$ of $NH_4^+$, we use the relation: $$ K_w = K_a \times K_b $$ where $K_w = 10^{-14}$

Therefore, we can find $K_a$ as follows:

K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-5}} = 10^{-9}

Next, we use the formula for calculating the concentration of hydrogen ions in a weak acid solution:

[\text{H}^+] = \sqrt{K_a \times C}

where $C$ is the concentration of $NH_4^+$ which is 0.02 M.

Substitute the values:

[\text{H}^+] = \sqrt{10^{-9} \times 0.02} = \sqrt{2 \times 10^{-11}} = \sqrt{2} \times 10^{-5.5}

Given $\log 2 = 0.301$, we calculate as follows:

\log(\sqrt{2}) = \frac{1}{2} \log 2 = \frac{0.301}{2} = 0.1505

So,

\log([\text{H}^+]) = 0.1505 - 5.5 = -5.3495

Converting to pH:

\text{pH} = - \log([\text{H}^+]) = 5.3495 \approx 5.35

Thus, the pH of a 0.02 M $NH_4Cl$ solution is approximately 5.35, which matches the correct answer.

Answer 2