Question

The $\left[H^{+}\right]$ of the resulting solution that is 0.01 M in acetic acid $(K_a=1.8\times {10}^{-5}M)$ and 0.01 M in benzoic acid. $(K_a=6.3\times {10}^{-5}M)$ will be:

• A. $4\times {10}^{-4}M$
• B. $16\times {10}^{-4}M$
• C. $81\times {10}^{-4}M$
• D. $9\times {10}^{-4}M$

Answer 1

The correct option is D $9\times {10}^{-4}M$

Let component 1 be acetic acid and component 2 be benzoic acid.
$C_1=C_2=0.01M{K}_{a1}=1.8\times {10}^{-5}M{K}_{a2}=6.3\times {10}^{-5}M$
For two weak acid solutions
$\left[H^{+}\right]=(K_{a1}{C}_1+K_{a2}{C}_2{)}^{0.5}[H^{+}]=[\left(0.01\right)(1.8\times {10}^{-5}+6.3\times {10}^{-5}){]}^{0.5}\left[H^{+}\right]=(8.1\times {10}^{-7}{)}^{0.5}[H^{+}]=9\times {10}^{-4}M$

Theory:
Weak acid 1 $\left(W{A}_1\right)$ + Weak acid 2 $\left(W{A}_2\right)$ :
Mixture of two weak monoprotic acids HA and HB with concentration $C_{1}and{C}_2$ and degree of dissociation ${\alpha }_{1}and{\alpha }_2$
For HA :
$HA\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+A^{-}\left(aq\right)$
$att=0C_{1}00$
$att=t_{eq}{C}_1-C_1{\alpha }_1(C_1{\alpha }_1+C_2{\alpha }_2)C_1{\alpha }_1$
For HB :
$HB\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+B^{-}\left(aq\right)$
$att=0C_{2}00$
$att=t_{eq}{C}_2-C_2{\alpha }_2(C_1{\alpha }_1+C_2{\alpha }_2)C_2{\alpha }_2$
Dissociation constant for HA $K_{a_1}$ :
$K_{a_1}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{C_1(1-{\alpha }_1)}$
Dissociation constant for HB $K_{a_2}$ :
$K_{a_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}{C_2(1-{\alpha }_2)}$
Since, ${\alpha }_{1}and{\alpha }_2$ are very small in comparison to unity for weak monoprotic acids. So $1-{\alpha }_1\approx 1and1-{\alpha }_2\approx 1$.
$C_1{K}_{a_1}+C_2{K}_{a_2}=(C_1{\alpha }_1+C_2{\alpha }_2{)}^2$
$\left[H^{+}\right]=C_1{\alpha }_1+C_2{\alpha }_2=\sqrt{C_1{K}_{a_1}+C_2{K}_{a_2}}$