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Solubility Product

The mass of s...

Question

The mass of substance containing $60 %$ $N a C l$ , $37 %$ $K C l$ , that should be weighed out for analysis so that, after the action of $25$ mL of $0.1 N A g N O_{3}$ solution, excess of ${A g}^{+}$ is back titrated with $5$ mL of $N H_{4} S C N$ solution, is__________.
Given that $1$ mL of $N H_{4} S C N = 1.1$ mL of $A g N O_{3}$ , write it in multiple of $100$ (only two digits).

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Solution

Suppose mass of sample used $= a g$
Thus, $w_{N a C l} = \frac{60 \times a}{100} = 0.6 a g$
$w_{K C l} = \frac{37 \times a}{100} = 0.37 a g$
This mixture reacts with $A g N O_{3}$ , the excess of $A g N O_{3}$ is back titrated by $N H_{4} S C N$ .
$∴$ Meq. of $A g N O_{3}$ added to mixture $= 25 \times 0.1 = 2.5$
The normaliy of $N H_{4} S C N$ can be derived by
Meq. of $N H_{4} S C N =$ Meq. of $A g N O_{3}$
$N \times 1 = 0.1 \times 1.1$
$∴ N_{N H_{4} S C N} = 0.11$
Meq. of $A g N O_{3}$ left = Meq. of $N H_{4} S C N = 5 \times N$
$= 5 \times 0.11 = 0.55$
Therefore, Meq. of $A g N O_{3}$ used for mixture
$= 2.5 - 0.55 = 1.95$
or Meq. of KCl+ Meq. of NaCl $= 1.95$
$\frac{0.37 a}{74.5} \times 1000 + \frac{0.6 a}{58.5} \times 1000 = 1.95$
$a = 0.128 = 013 g$
So, answer is $13$ .

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