$200 \mathrm{~mL}$ of $0.01 \,\mathrm{M} \,\mathrm{HCl}$ is mixed with $400 \mathrm{~mL}$ of $0.01 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}$. The $\mathrm{pH}$ of the mixture is _________.
Given: $\log {2}=0.30, \log 3=0.48, \log 5=0.70, \log 7=0.84, \log 11=1.04$
Solution
$$\begin{aligned} {\left[\mathrm{H}^{+}\right] } &=\frac{0.01 \times 200+2 \times 0.01 \times 400}{600} \\ &=\frac{0.01+2 \times 0.01 \times 2}{3} \\ &=\frac{0.01+0.04}{3} \\ &=\frac{5}{3} \times 10^{-2} \\ \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\ &=-\log \left(\frac{5}{3} \times 10^{-2}\right) \\ &=-\left[\log \frac{5}{3}+\log 10^{-2}\right] \\ &=-[\log 5-\log 3-2] \\ &=-0.7+0.48+2 \\ &=2.48-0.7 \\ &=1.78 \end{aligned}$$
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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