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Question

What will be the pH of a solution obtained by mixing 800mL of 0.05N sodium hydroxide and 200mL of 0.1N HCl, assuming the complete ionisation of the acid and the base?

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Solution

Number of milli-equivalents of NaOH=800×0.05=40
Number of milli-equivalents of HCl=200×0.1=20
Number of milli-equivalents of NaOH left after the addition of HCl =(4020)=20
Total volume =(200+800)mL=1000mL=1litre
20 milli-equivalents or 0.02 equivalents of NaOH are present in one litre
0.02N NaOH=0.02M NaOH (Mono-acidic) and the base is completely ionised
So, [OH]=0.02M
or [OH]=2×102M
pOH=log2×102=1.7
We know, pH+pOH=14
So, pH=(141.7)=12.3

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