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Q.
$100\, ml$ of a mixture of $NaOH$ and $Na_{2}SO_{4}$ is neutralised by 10 ml of 0.5 M $H_{2}SO_{4}$. Hence NaOH in $100$ mL solution is
Redox Reactions
Solution:
As $Na_{2}SO_{4}$ is already neutral so only $NaOH$ will be neutralized by $H_{2}SO_{4}$
$\therefore Eq. of NaOH = Eq. ofH_{2}SO_{4}$
$\frac{w\times1000}{40\times100} \times1\times \frac{100}{1000}=10\times10^{-3}\times0.5 \times2$
$ W=40 \times10^{-2} =0.4 g$