At $27^{\circ} \mathrm{C} .100 \mathrm{~mL}$ of 0.4 M HCl is mixed with 100 mL of 0.5 M NaOH solution. To the resultant solution. 800 mL of distilled water is added. What is the pH of final solution?

Moles of $\mathrm{HCl}=0.4 \mathrm{~m} \times 0.1 \mathrm{~L}=0.04$ moles For $\mathrm{NaOH}:-$ Moles of $\mathrm{NaOH}=0.5 \mathrm{M} \times 0.1 \mathrm{~L}=0.05$ moles HCl and NaOH react in $1: 1$ Ratio $\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}$ NaOH is in excess since 0.05 moles of NaOH react with 0.04 moles of HCl $(0.05$ moles -0.04 moles $)=0.01$ moles of NaOH remaining The total volume after mixing -100 ml of $\mathrm{HCl}+100 \mathrm{ml}$ of $\mathrm{NaOH}+800 \mathrm{ml}$ of distilled water $\Rightarrow 1000 \mathrm{ml}$ or 1 L so the concentration of $\mathrm{NaOH}=\frac{0.01 \mathrm{moles}}{1 \mathrm{I}}=0.01 \mathrm{~m}$ $\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]$ $\mathrm{pOH}=-\log [0.01]=2$ As we know pH + pOH=14 $\mathrm{pH}=14-\mathrm{pOH}=14-2=12$