Question

Question: An aqueous solution contains a mixture of 0.0500 M HCOOH (Ka = 1.77×10^-4) and 0.150 M CH₃CH₂COOH (Ka = 1.34×10^-5). Calculate the pH of this solution.

Short answer

Answer

pH for the mixture of given acids is 2.35.

Step-by-step solution

Calculating [H+] for both Acids 

For weak acids

$\left[H^{+}\right]=\sqrt{K_a\times c}$

So, for Formic acid,

$$\begin{gathered} \left[H^{+}\right]=\sqrt{K_a\times c} \\ \left[H^{+}\right]=\sqrt{177\times {10}^{-4}\times 0.0500} \\ \left[H^{+}\right]=3\times {10}^{-3} \end{gathered}$$

Similarly, for Acetic acid,

$$\begin{gathered} \left[H^{+}\right]=\sqrt{K_a\times c} \\ \left[H^{+}\right]=\sqrt{1.34\times {10}^{-5}\times 0.15} \\ \left[H^{+}\right]=1.42\times {10}^{-3} \end{gathered}$$

Calculating pH

Toal H⁺ ion concentration,

.

$$\begin{gathered} \left[H^{+}\right]=3\times {10}^{-3}+1.42\times {10}^{-3} \\ \left[H^{+}\right]=4.42\times {10}^{-3} \\ \text{pH}\text{=}\text{- log}\left[H^{+}\right] \\ \text{pH}\text{=}\text{- log}\left(4.42\times {10}^{-3}\right) \\ \text{pH}\text{=}\text{2.35} \end{gathered}$$

Hence, pH for the mixture of given acids is 2.35