1. Tardigrade
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  3. Chemistry
  4. 50 mL of 0.1 M HCl and 50 mL of 0.2 M NaOH are mixed. The pH of the resulting solution is

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Q. $50\, mL$ of $0.1\, M\,HCl$ and $50\, mL$ of $0.2\, M\, NaOH$ are mixed. The $pH$ of the resulting solution is

J & K CETJ & K CET 2004

Solution:

$50\, ML$ of $0.1\, M\,HCl =\frac{0.1 \times 50}{1000}=5 \times 10^{-3} 50\, ML$ of
$0.2\, M\,NaOH =\frac{0.2 \times 50}{100}=10 \times 10^{-3}$
Hence, after neutralisation $NaOH$ is left.
$=10 \times 10^{-3}-5 \times 10^{-3}=5 \times 10^{-3}$
Total volume $=100\, cc$ The concentration of
$NaOH =\frac{5 \times 10^{-3} \times 1000}{100}=0.05\, M$
$\left[ OH ^{-}\right]=0.05\, M =5 \times 10^{-2} M$
$pOH =-\log \left[ OH ^{-}\right]=-\log \left[5 \times 10^{-2}\right]$
$=1.3010 pH +p OH =14$
$p H =14-1.3010=12.699=12.70$