For the diprotic weak acid H2A, Ka1 = 3.8 × 10-6 and Ka2 = 5.8 × 10-9. What is the pH of a 0.0700 M solution of H2A? What are the equilibrium concentrations of H2A and A2- in this solution?
H2A = 0.0700M
Ka1= 3.8x10^-6
H2A -------------- HA- + H+
0.0700 0 0
-x +x +x
0.0700-x +x +x
Ka1= [HA-][H+]/[H2A]
3.8x10^-6 = x*x/(0.0700-x)
for solving the equation
x= 0.00051
[H+] = 0.00051M
-log[H+]= -log(0.00051)
PH= 3.29
[H2A] = 0.0700 - 0.00051=0.0695M
[HA-] = 0.00051M
HA- --------------- H+ + A-2
0.00051 0.00051
Ka2= [H+][A-2]/[HA-]
5.8x10^-9= 0.00051x[A-2]/0.00051
[A-2]= 5.8x10^-9M
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