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Q. What is the $pH$ of a solution obtained by mixing $10\,mL$ of $0.1\,M\,HCl$ and $40\,mL$ of $0.2\, M \,H_2SO_4$

Equilibrium

Solution:

Millimoles of $H^+$ from $HCl = 0.1\times 10=1$
Millimoles of $H^+$ from $H_2SO_4 = 0.2 \times 40 \times 2 = 16$
Conc, of $H^{+}=\frac{\text{Millimoles}}{\text{Volume}}=\frac{16+1}{40+10}$
$=\frac{17}{50}=0.34$
$pH=-log \left[H^{+}\right]$
$=-log\,\left(0.34\right)=0.468$