(Solved): A 100.0mL solution containing 0.976g of maleic acid (MW=116.072g/mol) is titrated with 0.366M ...

A $100.0\mathrm{mL}$ solution containing $0.976\mathrm{g}$ of maleic acid $(\mathrm{MW}=116.072\mathrm{g}/\mathrm{mol})$ is titrated with $0.366\mathrm{M}\mathrm{KOH}$. Calculate the $\mathrm{pH}$ of the solution after the addition of $46.0\mathrm{mL}$ of the $\mathrm{KOH}$ solution. Maleic acid has $\mathrm{p}{K}_{\mathrm{a}}$ values of 1.92 and 6.27. $\mathrm{pH}=$ At this $\mathrm{pH}$, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated as ${\mathrm{H}}_2\mathrm{M},{\mathrm{HM}}^{?}$, and ${\mathrm{M}}^{2?}$, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. $\left[{\mathrm{M}}^{2?}\right]$ M $\left[{\mathrm{HM}}^{?}\right]$ M $\left[{\mathrm{H}}_2\mathrm{M}\right]$ M