Consider (100.0 mathrm{~mL}) of a solution of (0.200 mathrm{M} mathrm{Na}_{2} mathrm{~A}), where (mathrm{A}^{2-}) is a base with

Question:

Consider $100.0 \mathrm{~mL}$ of a solution of $0.200 \mathrm{M} \mathrm{Na}_{2} \mathrm{~A}$, where $\mathrm{A}^{2-}$ is a base with corresponding acids $\mathrm{H}_{2} \mathrm{~A}$ $\left(K_{\mathrm{a}}=1.0 \times 10^{-3}ight)$ and $\mathrm{HA}^{-}\left(K_{\mathrm{a}}=1.0 \times 10^{-8}ight)$.

a. How much $1.00 \mathrm{M} \mathrm{HCl}$ must be added to this solution to reach $\mathrm{pH}=8.00$ ?

b. Calculate the $\mathrm{pH}$ at the second stoichiometric point of the titration of $0.200 \mathrm{M} \mathrm{Na}_{2} \mathrm{~A}$, with $1.00 \mathrm{M} \mathrm{HCl}$.

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