A solution is prepared by adding $1\mathrm{.}0$ mol of solid lithium hydroxide to $2\mathrm{.}0$ L of $0\mathrm{.}5$ M weak acid $($HA$).$ The pKa of the weak acid is pKa$($HA$).$ What is the pH of the solution?

pH $=$ pKa $($HA$)+0\mathrm{.}31$

Incorrect Response

pH $=$ pKa$($HA$)0\mathrm{.}31$

pH $<$ pKa$($HA$)0\mathrm{.}31$

Correct Answer

pH $>$ pKa$($HA$)+0\mathrm{.}31$