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Question

What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?

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Solution

One mole of NaOH is completely neutralised by one mole of HCl.

Hence, 0.01 mole of NaOH will be completely neutralised by 0.01 mole of HCl.

NaOH left unneutralised = 0.1 - 0.01 mol

= 0.09 mol

As equal volumes of two solutions are mixed,

$[\mathrm{OH}{]}^{-}=\frac{0.09}{2}=0.045\mathrm{M}$

$\Rightarrow \mathrm{pOH}=-\mathrm{log}(0.045)=1.35$

$\therefore \mathrm{pH}=14-1.35=12.65$

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