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Reaction with Metals

25 ml of a so...

Question

25 ml of a solution containing $H C l$ and sulphuric acid required 10 ml of a 2.0 N $N a O H$ solution for neutralisation. 10 ml of the same acid mixture on being treated with an excess of $B a C l_{2}$ gave 0.699 gm of $B a S O_{4}$ . The normality of $H C l$ is .

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Solution

$X_{1}$ mmol of HCl and $X_{2}$ mmol of $H_{2} S O_{4}$ are present in 25 ml acid mixture.
So, $X_{1} + 2 X_{2} = 10 \times 2$ (1)
Again,
$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$
Since we are taking only 10 ml of the acidic solution,
$\frac{X_{2} \times 10}{25} = \frac{0.699}{233}$
$\Rightarrow X_{2} = 7.5 m m o l$
$\Rightarrow X_{1} = 5 m m o l$
So, normality of $H C l = \frac{5}{25} = 0.2 N$

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