Question

Calculate the $pH$ of solution obtained by mixing $10mL$ of $0.2MHCl$ and $40mL$ of $0.1M{H}_{2}S{O}_4$.

Answer 1

$H_{2}S{O}_4$ dissociates:

$H_{2}S{O}_4\to H^{+}+HS{O}^{4-}$

This first dissociation step occurs almost completely which makes $H_{2}S{O}_4$ a strong acid

second step:

$HS{O}_4^{-}\to H^{+}+S{O}_4^{2-}$ occurs to a small extent $HS{O}_4^{-}$ is a weak acid . Its contrbution to the $H^{+}$ will be small - and even further reduced in the presence of the high $H^{+}$ from the $HCl$ .

HCl is a strong acid that dissociuates completely.

The total $H^{+}$ in the final solution is calculated:

Mol $H_{2}S{O}_4$$in40mLof0.2Msolution=$\frac{40}{100}\times 0.2=0.008$ mol H2SO4

This will dissociate to produce 0.008 mol $H^{+}$

Mol HCl in 10mL of 0.1M solution $\frac{10}{10}\times 0.1=0.001$ mol of $HCl$

This will dissociate to produce 0.001 mol $H^{+}$

There is therefore a total of 0.009 mol $H^{+}$ dissolved in 50mL solution = 0.050L

Molarity of $H^{+}=\frac{0.0009}{0.050}=0.18$ M

pH = -log [H+]

pH = -log 0.18

pH = 0.74