Respuesta :
Answer:
Following are the solution to the given question:
Explanation:
Let us indicate H3AsO4, an H3A triprotic acid; the ionising equations are as follows:
[tex]\to H_3A \Leftrightarrow H^+ + H_2A^- \\\\[/tex]
[tex]\to Ka_1 = \frac{[H^+][H_2A^-]}{[H_3A]}[/tex]
[tex]= 10-2.2 = 6.31 \times 10^{-3} \ \ \ \ \ \ \ \ [since\ pKa_1 = 2.2\ and \ Ka = -\log_{10}(pKa)]\\[/tex]
[tex]\to H_2A^{-} \Leftrightarrow H^{+} + HA_2^{-}; \\[/tex]
[tex]\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)[/tex]
The initial pH; we've 0.00 mL of 0.100 M NaOH – H3A is the main species; Ka1>>Ka2>>Ka3 is also noted. They therefore ignore H2A's disconnection but set up next ICE chart.
[tex]\to H_3A \Leftrightarrow H^+ + H_2A^- \\\\[/tex]
[tex]initial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\\\[/tex]
[tex]change \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + x\\\\equilibrium \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (0.100 - x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\\\\[/tex]
[tex]Ka_1 = \frac{[H^+][H_2A^-]}{[H_3A]} = \frac{(x)(x)}{(0.100 - x)}[/tex]
[tex]\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[/tex]
