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  4. By adding 20 ml 0.1 N HCl to 20 ml 0.001 N KOH, the pH of the obtained solution will be

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Q. By adding $20\, ml \,0.1 \,N HCl$ to $20\, ml \,0.001 \,N\, KOH$, the $pH$ of the obtained solution will be

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Solution:

Milliequivalents of acid $=N_{1}V_{1}=20 \times 0.1\,N=2$
Milliequivalents of base $=N_{2}V_{2}=20\times 0.001\,N$
$=0.02$
Remaining acid $= 2 - 0.02 = 1.8$
$[H^{+}]=\frac{1.8}{V_{1}+V_{2}}$
$=\frac{1.8}{40}$
$=0.0045\,M$
$PH=-log[H^{+}]=1.35$