Posted by Geroge B. on Saturday, April 5, 2008 at 6:37pm.
Let's call Lactic acid HL.
HL ==> H^+ + L^-
Ka = (H^+)(L^-)/(HL) = 1.4 x 10^-4
initially before ionization:
(HL) = 0.001 M
(H^+) = 0
(L^-) = 0
change:
(H^+) = +x
)L^-) = +x
(HL) = 0.001 - x
equilibrium:
(H^+) = +x
(L^-) = +x
(HL) = 0.001 - x
Substitute the equilibrium values shown into the Ka expression and solve for x.
Then convert (H^+) to pH by pH = -log(H^+)
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