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$25 \mathrm{~mL}, 0.2 \mathrm{MCa}(\mathrm{OH})_2$ is neutralised by $10 \mathrm{~mL}$ of $1 \mathrm{MHCl}$. Then $\mathrm{pH}$ of resulting solution is
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The correct answer is:
7
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 $=7$.
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 $=7$.
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