A 0.100M solution of a monoprotic weak acid has a pH of 3.00. What is the pKa of this acid?
PH= 3.00
-log[H+] = 3.00
[H+] = 10^-3.00
[H+]= 10^-3.00 M
Concentration of mono protic acid = 0.100M
for weak acdis
[H+] = square root of KaxC
[H+]^2 = KaxC
Ka = [H+]^2/C = (10^-3)^2/0.1 = 10^-6 x10 = 1.0x10^-5
Ka= 1.0x10^-5
-log(Ka) = -log(1.0x10^-5)
PKa = 5.0
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