Question

Determine the $pH$ value of a solution obtained by mixing $25mL$ of $0.2MHCl$ and $50mL$ of $0.25NNaOH$ solutions.

Answer 1

$HCl\to 25ml$ of $0.2M$

$NaOH\to 50ml$ of $0.25M$

no. of $m$ moles of $HCl=25\times 0.2=5$ mmole

no. of $m$ moles of $NaOH=50\times \frac{25}{100}=12.5$ mmole

$NaOH+HCl⟶NaCl+H_{2}O$

Initially $12.55$
Finally $12.5-50$

$\therefore$ mmoles of $NaOH$ remained $=7.5$

Volume of final solution = Volume of $HCl+$ volume of $NaOH=25+50=75ml$

$\therefore$ Molarity of $NaOH=\frac{75}{750}=0.1$

$\left[O{H}^{-}\right]=0.1M$

$pOH=-\log \left[O{H}^{-}\right]=1\therefore pH=13$