How many mL (nearest integer) of $0.30 M C a C l_{2}$ must be added to $400$ mL of $0.20 M K C l$ to produce a solution of $0.25 M C l^{-}$ ?

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Solution

$0.30 M C a C l_{2}$ on dissociation will give $0.60 M$ chloride ion solution.
Let $V_{1}$ mL be the volume of the $C a C l_{2}$ solution.
$0.20 M$ $K C l$ solution on dissociation will give $0.20 M$ chloride ion solution.
The volume of $K C l$ solution is $400$ mL.
The resultant molarity of chloride ion solution is $0.25 M$ .

It is given by the expression $\frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}}$ .

Hence, $0.25 = \frac{0.60 V_{1} + 0.20 \times 400}{V_{1} + 400}$ .

Hence, $V_{1} = 57.1 = 57$ mL

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