Question

Equal volumes of three acid solutions of pH $3, 4$ and $5$ are mixed in a vessel. What will be the $ H^+ $ ion concentration in the mixture?

• A. $ 1.11 \times 10^{-4}M $
• B. $ 3.7 \times 10^{-4}M $
• C. $ 3.7 \times 10^{-3}M $
• D. $ 1.11 \times 10^{-3}M $

Answer 1

Answer: (B)

Let the volume of each acid $=V$
$pH$ of first, second and third acids $=3,4$ and $5$ respectively
$\left[H^{+}\right]$of first acid $\left(M_{1}\right)=1 \times 10^{-3}\left[\therefore H^{+}=1 \times 10^{-p H}\right]$
$\left[H^{+}\right]$of second acid $\left(M_{2}\right)=1 \times 10^{-4}$
$\left[H^{+}\right]$of third acid $\left(M_{3}\right)=1 \times 10^{-5}$
Total $\left[ H ^{+}\right]$concentrated of mixture
$(M)=\frac{M_{1} V_{1}+M_{2} V_{2}+M_{3} V_{3}}{V_{1}+V_{2}+V_{3}} $
$=\frac{1 \times 10^{-3} \times V+1 \times 10^{-4} \times V+1 \times 10^{-5} \times V}{V+V+V} $
$=\frac{1 \times 10^{-3} \times V(1+0.1+0.01)}{3 V}$
$=\frac{1.11 \times 10^{-3}}{3}=3.7 \times 10^{-4} M$