$25 \mathrm{~mL}, 0.2 \mathrm{MCa}(\mathrm{OH})_2$ is neutralised by $10 \mathrm{~mL}$ of $1 \mathrm{MHCl}$. Then $\mathrm{pH}$ of resulting solution is

Number of millimoles of base (i.e., $\left.\mathrm{Ca}(\mathrm{OH})_2\right)$ $=N_1 V_1=2 \times M_1 \times V_1=2 \times 0.2 \times 25=10$ Number of millimoles of acid (i.e., $\mathrm{HCl})=\mathrm{N}_2 V_2$ $$ =10 \times 1=10 $$ As, no. of millimoles of acid $=$ no. of millimoles of base $\therefore$ Acid is completely neutralised by base forming a neutral solution. $\therefore \mathrm{pH}$ of the resulting solution $=7$.