Chemistry, 11.01.2021 15:20 morganluvsblueow9wtm
) Calculate the pH of a polyprotic acid given and sketch the titration curves for the following reactions: (a) A 20.0-mL aliquot of 0.100M arsenic acid, H3AsO4, with 0.100M NaOH
Answers: 2
![\to Ka_1 = \frac{[H^+][H_2A^-]}{[H_3A]}](/tpl/images/1025/9491/6962a.png)
![= 10-2.2 = 6.31 \times 10^{-3} \ \ \ \ \ \ \ \ [since\ pKa_1 = 2.2\ and \ Ka = -\log_{10}(pKa)]\\](/tpl/images/1025/9491/eec45.png)
![\to Ka_2 = \frac{[H^+][HA_2^-]}{[H_2A^-]} = 10-6.8 = 1.58 \times 10^{-7} \\(since pKa2 = 6.8)\\\\\to HA_2^- \Leftrightarrow H^+ + A_3^- ;\\\\ \to Ka_3 = \frac{[H^+][A_3^-]}{[HA_2^-]} = 10-11.6 = 2.51 \times 10^{-12} \\ (given pKa_3 = 11.6)](/tpl/images/1025/9491/119c9.png)
![Ka_1 = \frac{[H^+][H_2A^-]}{[H_3A]} = \frac{(x)(x)}{(0.100 - x)}](/tpl/images/1025/9491/3211c.png)
![\text{x to be much smaller than 0.100 M}\\\\to 6.31 \times 10^{-3} = \frac{x^2}{0.100}\\\\\to x^2 = 6.31 \times 10^{-4}\\\\ \to x = 0.025\ M\\\\\to pH = -\log_{10}[H^+] = -\log_{10}(0.025) = 1.60](/tpl/images/1025/9491/7bd66.png)
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) Calculate the pH of a polyprotic acid given and sketch the titration curves for the following reac...
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