Answer to Question #105040 in Organic Chemistry for Anthony

Question #105040
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?
1
Expert's answer
2020-03-10T09:08:43-0400

"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"



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