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Normality

50 mL of 1 M ...

Question

50 mL of 1 M HCl, 100 mL of 0.5 M $H N O_{3},$ and x mL of 5 M $H_{2} S O_{4}$ are mixed together and the total volume is made upto 1.0 L with water. 100 mL of this solution exactly neutralises 10 mL of M/3 $A l_{2} (C O_{3})_{3} .$ Calculate the value of x:___

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Solution

Total mEq of acid = $50 \times 1 + 100 \times 0.5 + x \times 5 \times 2$ (n factor)
= (100 + 10x)mEq
$= \frac{(100 + 10 x)}{100 m L} N$
$N_{1} V_{1} (A c i d) = N_{1} V_{1} [A l_{2} (C O_{3}^{2 -})_{3}]$ (Total charge = 6)
(n = 6)
$∴ \frac{(100 + 10 x)}{100 m L} N \times 100 m L = 10 m L \times \frac{1}{3} \times 6$
(100 + 10x) = 200
$∴$ x = 10 mL

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