$100 \mathrm{~cm}^{3}$ of $0.1 \mathrm{NHCl}$ is mixed with $100 \mathrm{~cm}^{3}$ of $0.2$ N NaOH solution.…

$100 \mathrm{~cm}^{3}$ of $0.1 \mathrm{NHCl}$ is mixed with $100 \mathrm{~cm}^{3}$ of $0.2$ N NaOH solution. The resulting solution is
  1. $0.1 \mathrm{~N}$ and the solution is basic
  2. $0.1 \mathrm{~N}$ and the solution is acidic
  3. $0.05 \mathrm{~N}$ and the solution is basic
  4. $0.05 \mathrm{~N}$ and the solution is acidic

Solution

Normality
$=\frac{\mathrm{N}_{1} \mathrm{~V}_{1}-\mathrm{N}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}+\mathrm{V}_{2}}=\frac{0.2 \times 100-0.1 \times 100}{100+100}$
$=\frac{10}{200}=0.05 \mathrm{~N} \mathrm{NaOH}$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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