(6 pts) Use the two calculation methods identified below, to calculate the pH of
ID: 884465 • Letter: #
Question
(6 pts) Use the two calculation methods identified below, to calculate the pH of a solution of 1.23 x 10^-3 M lactic acid. Ka constant info on final page. Carry out each pH calculation to +-0.001 pH unit in order to determine the differences in calculation techniques. In other words, neglect the significant figure convention for this problem. Please show an ICE table setup - with the equation showing the equilibrium, and how you plugged values into your Ka expression. (a)Assume the acid dissociation is negligible compared to CHA. In other words, assume that CHA = [HA], or that the concentration of the acid before and after equilibrium is established doesn't change very much. (Please show where you make this assumption in your work.)Explanation / Answer
lactic acid formula C3H6O3 = C3H5O3H
C3H5O3H (lactic acid ) ---------------------------------> C3H5O3- (lactate) + H+
1.23 x 10^-3 0 0 -----------------------> initial
-x + x +x ----------------------> changed
1.23 x 10^-3 -x x x -------------------------> equilibrium
Ka = [C3H5O3-][H+]/[C3H5O3H]
Ka = (x)(x) / (1.23 x 10^-3 -x )
1.4 x 10^-4 = x^2 / (1.23 x 10^-3 -x )
x can be neglected in denominator
x^2 = 1.4 x 10^-4 x 1.23 x 10^-3
x^2 = 1.722 x 10^-7
x = 4.15 x 10^-4
x = [H+] = 4.15 x 10^-4 M
pH = -log [H+]
pH = -log (4.15 x 10^-4)
pH = 3.38
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