Question

The ratio in which $0.2MNaCl$ and $0.1MCa{Cl}_2$ 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 (write $\frac{x}{y}$ as $xy$)

Answer 1

Suppose $V_2$ mL of $NaCl$ is mixed with $V_1$ mL of $Ca{Cl}_2$ solution.
$\therefore$ Millimoles of $Ca{Cl}_2$ mixed $=$ $0.2\times V_1$
Millimoles of $NaCl$ mixed $=$ $0.1\times V_2$

$\therefore$ Molarity of ${Cl}^{-}$ in mixture $=$ $\left[{Cl}^{-}\right]$ of $NaCl$ $+\left[{Cl}^{-}\right]$ of $Ca{Cl}_2$

$=\left[\frac{0.2\times V_1}{V_1+V_2}\right]+\left[\frac{0.1\times 2\times V_2}{V_1+V_2}\right]$ ($\because$ $1$ mole of $Ca{Cl}_2$ gives $2$ moles ${Cl}^{-}$)

$=\frac{[0.2V_1+0.2V_2]}{[V_1+V_2]}$

Similarly, molarity of ${Na}^{+}$ and ${Ca}^{2+}$ in mixture

$=\left[{Na}^{+}\right]$ of $NaCl+$ [${Ca}^{2+}$] of $Ca{Cl}_2$

$=\left[\frac{0.2\times V_1}{V_1+V_2}\right]+\left[\frac{0.1\times V_2}{V_1+V_2}\right]=\left[\frac{0.2V_1\times 0.1V_2}{V_1+V_2}\right]$

Now, given molarity of ${Cl}^{-}=\frac{3}{2}\times$ molarity of ${Na}^{+}$ and ${Ca}^{2+}$

$\frac{0.2V_1+0.2V_2}{V_1+V_2}=\frac{3}{2}\left[\frac{0.2V_1\times 0.1V_2}{V_1+V_2}\right]$

$\frac{V_1}{V_2}=\frac{1}{2}$

So, the answer is $12$.