Question

Calculate the $\left[{Cl}^{-}\right],\left[{Na}^{+}\right],\left[H^{+}\right],\left[{OH}^{-}\right]$ and pH of the resulting solution obtained by mixing $50mL$ of $0.6N$ $HCl$ and $50mL$ of $0.3N$ $NaOH$.

Answer 1

no. of moles of $HCl$ is $50\times 0.6=30$

no of moles of $NaOH$ is $50\times 0.3=15$

$HCl+NaOH⟶NaCl+H_{2}O$

Milli-equivalents $50\times 0.6$ $50\times 0.3$ $0$ $0$

before reaction $=30$ $=15$

Milli-equivalents $(30-15)$

after reaction $=15$ $0$ $15$ $15$

total volume $=50+50=100mL$

$\left[{Cl}^{-}\right]=\frac{15+15}{100}=0.3M$

$\left[{Na}^{+}\right]=\frac{15}{100}=0.15M$

$\left[H^{+}\right]=\frac{15}{100}=0.15M$

$\left[{OH}^{-}\right]=\frac{{10}^{-14}}{0.15}=6.6\times {10}^{-14}$

$pH=-\log \left[H^{+}\right]=-\log 0.15=0.8239$