Question

) 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

Answer 1

Following are the solution to the given question:

Step-by-step explanation:

Let us indicate H3AsO4, an H3A triprotic acid; the ionising equations are as follows:

`\to H_3A \Leftrightarrow H^+ + H_2A^- \\\\`

`\to Ka_1 = ([H^+][H_2A^-])/([H_3A])`

`= 10-2.2 = 6.31 * 10^(-3) \ \ \ \ \ \ \ \ [since\ pKa_1 = 2.2\ and \ Ka = -\log_(10)(pKa)]\\`

`\to H_2A^(-) \Leftrightarrow H^(+) + HA_2^(-); \\`

`\to Ka_2 = ([H^+][HA_2^-])/([H_2A^-]) = 10-6.8 = 1.58 * 10^(-7) \\(since pKa2 = 6.8)\\\\\to HA_2^- \Leftrightarrow H^+ + A_3^- ;\\\\ \to Ka_3 = ([H^+][A_3^-])/([HA_2^-]) = 10-11.6 = 2.51 * 10^(-12) \\ (given pKa_3 = 11.6)`

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.

`\to H_3A \Leftrightarrow H^+ + H_2A^- \\\\`

`initial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\\\`

`change \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + x\\\\equilibrium \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (0.100 - x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\\\\`

`Ka_1 = ([H^+][H_2A^-])/([H_3A]) = ((x)(x))/((0.100 - x))`

`\text{x to be much smaller than 0.100 M}\\\\to 6.31 * 10^(-3) = (x^2)/(0.100)\\\\\to x^2 = 6.31 * 10^(-4)\\\\ \to x = 0.025\ M\\\\\to pH = -\log_(10)[H^+] = -\log_(10)(0.025) = 1.60`