Q. The molarity of $OH^{-}$ ion in the final solution, if $200ml$ of $8MHCl$ solution is mixed with $800ml$ of $2MCa\left(\right.OH\left(\left.\right)_{2}$ solution, is

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A

$0.08M$

B

$1.6M$

C

$3.2M$

D

$0.16M$

Solution:

$2HCl+Ca\left(\right.OH\left(\left.\right)_{2} \rightarrow \left(CaCl\right)_{2}+2H_{2}O$
initially $8\times \frac{200}{1000}$ $2\times \frac{800}{1000}$ $-$ $-$
moles $=1.6$ $=1.6$
Finally $1.6-\frac{1 . 6}{2}\Rightarrow 0.08$
$\left[\left(OH\right)^{-}\right]=\frac{\text{ moles of } \left(OH\right)^{-}}{\text{Volume of solution( } L \left.\right)}=\frac{2 \times 0 . 08}{1}=1.6M$ Ans.

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