The ratio in which $0.2$ M $N a C l$ and $0.1$ $C a {C l}_{2}$ solutions are to be mixed so that in the resulting solution the concentration of negative ions is $50 %$ greater than the concentration of positive ions, is $\frac{x}{y}$ . Then, $x + y$ is:

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Solution

Suppose $V_{2} m L$ of $N a C l$ is mixed with $V_{1}$ mL of $C a {C l}_{2}$ solution.
$∴$ Milli mole of $C a {C l}_{2}$ mixed $=$ $0.2 \times V_{1}$
Milli mole of $N a C l$ mixed $=$ $0.1 \times V_{2}$
$∴$ Molarity of ${C l}^{-}$ in mixture $=$ $[{C l}^{-}]$ of $N a C l$ + $[{C l}^{-}]$ of $C a {C l}_{2}$
$=$ $[\frac{0.2 \times V_{1}}{V_{1} + V_{2}}] + [\frac{0.1 \times 2 \times V_{2}}{V_{1} + V_{2}}]$
( $∵$ 1 mole of $C a {C l}_{2}$ gives 2 mole ${C l}^{-}$ )
$=$ $\frac{[0.2 V_{1} + 0.2 V_{2}]}{[V_{1} + V_{2}]}$
Similarly, molarity of ${N a}^{+}$ and ${C a}^{2 +}$ in mixture
$=$ [ ${N a}^{+}$ ] of $N a C l$ + [ ${C a}^{2 +}$ ] of $C a {C l}_{2}$
$=$ $[\frac{0.2 \times V_{1}}{V_{1} + V_{2}}] + [\frac{0.1 \times V_{2}}{V_{1} + V_{2}}] = [\frac{0.2 V_{1} \times 0.1 V_{2}}{V_{1} + V_{2}}]$
Now, given molarity of ${C l}^{-} = 3 / 2 \times m o l a r i t y o f {N a}^{+} a n d {C a}^{2 +}$
$\frac{0.2 V_{1} + 0.2 V_{2}}{V_{1} + V_{2}} = \frac{3}{2} [\frac{0.2 V_{1} \times 0.1 V_{2}}{V_{1} + V_{2}}]$
$\frac{V_{1}}{V_{2}} = \frac{1}{2}$
So, answer is $1 + 2 = 3$ .

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