When a mixtur...

Question

When a mixture of $N a B r$ and $N a C l$ is repeatedly digested with $H_{2} S O_{4}$ , all the halogens are expelled and ${N a}_{2} S O_{4}$ is formed quantitatively. With a particular mixture, it was found that the mass of ${N a}_{2} S O_{4}$ obtained is precisely the same as the mass of $N a B r$ and $N a C l$ mixture taken. The mass ratio of $N a C l$ and $N a B r$ in the mixture is (as nearest integer only) :

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Solution

$N a B r H_{2} S O_{4} - --- \to {N a}_{2} S O_{4}$
$a$ g
$N a C l H_{2} S O_{4} - --- \to {N a}_{2} S O_{4}$
$b$ g
$∴$ Meq. of $N a B r +$ Meq. of $N a C l$ $=$ Meq. of ${N a}_{2} S O_{4}$
Meq. of ${N a}_{2} S O_{4}$ formed $= \frac{a}{103} \times 1000 + \frac{b}{58.5} \times 1000$ $= 9.71 a + 17.09 b$
$\frac{w}{\frac{142}{2}} \times 1000 = 9.71 a + 17.09$
$w = \frac{142}{2000} [9.71 a + 17.09 b]$
Also,
mass of $N a C l +$ mass of $N a B r$ $=$ mass of ${N a}_{2} S O_{4}$ formed
$a + b = \frac{142}{2000} [9.71 a + 17.09 b]$

$\frac{a}{b} = \frac{426.78}{621.18} = 0.687$

Ratio of $\frac{NaCl}{NaBr}$ $= \frac{b}{a} = \frac{1.455}{1}$
So, the ratio as nearest integer is $1$ .

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