In the titration of (100.0 mathrm{~mL}) of a (0.0500 mathrm{M}) solution of acid (mathrm{H}_{3} mathrm{~A}left(K_{mathrm{a}_{1}}=1.0 times 10^{-3},

Question:

In the titration of $100.0 \mathrm{~mL}$ of a $0.0500 \mathrm{M}$ solution of acid $\mathrm{H}_{3} \mathrm{~A}\left(K_{\mathrm{a}_{1}}=1.0 \times 10^{-3}, K_{\mathrm{a}_{2}}=5.0 \times 10^{-8}, K_{\mathrm{a}_{3}}=ight.$ $2.0 \times 10^{-12}$ ), calculate the volume of $1.00 \mathrm{M} \mathrm{NaOH}$ required to reach $\mathrm{pH}$ values of 9.50 and 4.00 .

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