What is the equilibrium concentration of H2PO4– in a 0.202 M solution of H3PO4(aq)? (Ka1 =...

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Question

What is the equilibrium concentration of H2PO4– in a 0.202 M solution of H3PO4(aq)? (Ka1 = 7.5 x 10^-3,Ka2 =6.2x10–8,Ka3 =4.8x10–13)

please list full step by step on how to solve problem.

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Answer 1

Answer #1

H3Po4(aq) ------------------> H2Po4^- (aq) + H^+ (aq)

I 0.202 0 0

C -x +x +x

E 0.202-x +x +x

ka1 = [H2PO4^-][H^+]/[H3PO4]

7.5*10^-3 = x*x/(0.202-x)

7.5*10^-3*(0.202-x) = x^2

x = 0.03535

[H2PO4^-] = x = 0.03535M

the equilibrium concentration of H2PO4– = 0.03535M

Answer 2

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