Question

Arsenic acid is a triprotic acid in water, with
Ka1= 5.5×10^-3
Ka2=1.7×10^-7
Ka3=5.1×10^-12

What is the pH of 0.80 M Na3AsO4?

Answer 1

$K_b =\frac{K_w}{K_a}$

$K_{b1}=\frac{1\times10^{-14}}{5.1\times10^{-12}}= 1.96\times10^{-3}$

$K_{b2} = \frac{1\times10^{-14}}{1.7\times10^{-7}}=5.88\times10^{-8}$

$K_{b3}=\frac{1\times10^{-14}}{5.5\times10^{-3}}=1.82\times10^{-12}$

1) $PO_4^{3-} + H_2O \leftrightarrow HPO_4^{2-}+OH^-$

I 0.8 0 0

C -x +x +x

E 0.8-x x x

$K_{b1} = \frac{[HPO4^{2-}][Oh^-]}{PO4^{3-}}=\frac{x^2}{0.8-x}=1.96\times10^{-3}$

$x=0.039$

$[OH^-]= 0.039M$

$[HPO_4^{3-}]=0.039M$

2) $HPO_4^{3-}+H_2O \leftrightarrow H_2PO_4^{2-}+ OH^-$

I 0.039 0 0

C -x +x +x

E 0.039-x x x

$K_{b2}=\frac{[H_2PO_4^{2-}][OH^-]}{[HPO_4^{3-}]}=\frac{x^2}{0.039-x}=5.88\times10^{-8}$

$x= 0.00024$

$[OH^-] = 0.00024M$

$[H_2PO_4^{2-}] = 0.00024 M$

3) $H_2PO_2^{2-} + H_2O \leftrightarrow H_3PO_4 + OH^-$

I 0.00024 0 0

C -x +x +x

E 0.00024-x x x

$K_{b3} = \frac{[H_3PO_4][OH^-]}{[H_2PO_4^{2-}]}=\frac{x^2}{0.00024-x}=1.82\times10^{-12}$

$x= 2.1\times 10^{-8}$

$[OH^-] = 2.1\times10^{-8}M$

$[OH^-]_{total} = 0.039+0.00024+2.1\times10^{-8} = 0.039$

$pOH = -log_{10}(0.039) = 1.4$

$pH = 14-pOH = 14-1.4 = 12.6$