Question

Two weak monobasic organic acids HA and HB have dissociation constants as $3.0\times {10}^{-5}$ and $1.5\times {10}^{-5}$, respectively, at ${25}^{\circ }$. If 500 mL of 1M solutions of each of these two acids are mixed to produce 1 L of mixed acid solution, what is the pH of the resulting solution?

• A. pH = 3.32
• B. pH = 4.32
• C. pH = 2.32
• D. pH = 5.23

Answer 1

The correct option is C

pH = 2.32

$\begin{array}{ccccc}\text{HA}& \rightleftharpoons & H^{+}& +& A^{-}\\ \text{C}& & & & \\ \text{C-x}& & \text{x+y}& & \text{x}\\ \text{HB}& \leftrightharpoons & {\text{H}}^{⨁}& +& {\text{B}}^{-}\\ \text{C}& & & & \\ \text{C-y}& & \text{x+y}& & \text{y}\end{array}$

$=K_{a\left(HA\right)}=\frac{(x+y)x}{C}$ and $K_{a\left(HB\right)}=\frac{(x+y)y}{C}$

Divide the two expression to get

$\frac{1}{2}=\frac{x}{y}\Rightarrow x=2y$

Substitute for $y=\frac{1}{2}x$ in $K_{a\left(HA\right)}=\frac{x^2+xy}{C}$

$\Rightarrow 3.0\times {10}^{-5}=\frac{x^2+0.5x^2}{0.5}\Rightarrow x=\sqrt{10}\times {10}^{-3}\text{M}$ and $y=\frac{\sqrt{10}}{2}\times {10}^{-3}\text{M}$

$\left[{\text{H}}^{⨁}\right]=\text{x+y}=\frac{3\sqrt{10}}{2}\times {10}^{-3}\text{M}$

pH $=-log(\frac{3\sqrt{10}}{2}\times {10}^{-3})\Rightarrow \text{pH}=2.32$

You don't actually have to calculate the last part, if you get the exponent right, you know that the pH has to be between 2 and 3. The only option that fits is 2.32!