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Q.
What shall be the $ pH $ of a solution formed by mixing $ 10 \,mL $ of $ 0.1 \,M \,H_{2}SO_{4} $ and $ 10 \,mL $ of $ \frac{N}{10} $ $ KOH $ ?
AMUAMU 2014Equilibrium
Solution:
Milliequivalent of $H^{+}=10 \times 0.1 \times 2=0$
Milliequivalent of $OH^{-} =10\times \frac{1}{10}=1$
Normality of solution $=\frac{2-1}{20}=\frac{1}{20}=0.05$
Molarity of the resulting solution
$=2\times 0.05=0.1$
$pH$ of the solution $=-log [H^{+}]=-log (0.1)$
$=-log(10^{-1})=1$