Calculate hydrogen ion concentrations of a buffer that is 0.05M (KHP) and 0.2M in (K2P), C6H4 (COOH)2 ka1=1.12*10-3, ka2=3.91x10-6?
- hcbiochemLv 71 year ago
The only Ka that is relevant is Ka2. Using the equilibrium constant expression for that ionization:
Ka2 = 3.91X10^-6 = [H+][P2-]/[HP-]
[H+](0.20) / 0.05 = 3.91X10^-6
[H+] = 9.8X10^-7 M
pH = 6.01
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