Find the pH of a 0.0150M H2SO4 solution (Ka=0.012)
H2SO4 is strong acid
H2So4 ----------> H^+ (aq) + HSO4^-
I 0.015 0 0
C -x +x +x
E 0.015-x +x +x
ka = [H^+][HSO4^-]/[H2SO4]
0.012 = x*x/0.015-x
0.012*(0.015-x) = x^2
x = 0.0087
[H^+] = x = 0.0087 M
PH = -log[H^+]
= -log0.0087
= 2.0604 >>>>>answer
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