1)
[H+] = 1.45*10^-3 M
use:
pH = -log [H+]
= -log (1.45*10^-3)
= 2.84
Answer: 2.84
2)
let the weak acid be HA
HA -----> H+ + A-
2.9*10^-2 0 0
2.9*10^-2-x x x
x = [H+] = 1.45*10^-3 M
% ionization = x*100/c
= (1.45*10^-3)*100 / 0.0290
= 5.0 %
Answer: 5.0 %
3)
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 1.45*10^-3*1.45*10^-3/(0.029-1.45*10^-3)
Ka = 7.63*10^-5
Answer: 7.63*10^-5
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