Posted by Sarah on Sunday, April 6, 2008 at 10:42pm.
Lactic acid, HC3H5O3(aq) is a weak acid that gives yougurt its sour taste(Yeeeeecccckkk). Calculate the pH of a 0.0010 mol/L solution of Lactic acid. The Ka for lactic acid is 1.4 x 10^4
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chemistry  DrBob222, Saturday, April 5, 2008 at 7:39pm
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^+)
Here is my work
Ka = H^+)(L^)/(HL) = 1.4 x 10^4
=(+x)(+x)/(0.001  x ) = 1.4 x 10^4
= (x^2) / (0.001  x) = 1.4 x 10^4
What is my next step sorry:)

chemistry  DrBob222, Sunday, April 6, 2008 at 10:53pm
You have two choices.
1)The simple choice, and one which doesn't always work, is to assume x is very small in comparison with 0.001 so that 0.001  x = 0.001. That way you don't have a quadratic equation. You can try that, then see if 0.001  x = about 0.001 or not. If x is too large, that won't work.
#2 choice. Solve the quadratic.
x^2 = 1.4 x 10^4 x (0.001  x) and go from there. Many caculators now have a quadratic solver built in (or they can be programmed to solve a quadratic).
#3 choice. Technically, this isn't another choice. But sometimes the quadratic can be solved by successive approximations which may or may not be easier than solving the quadratic by brute force. I don't know if this one will work that way or not.
I'll leave you with those choices and you let me know which ones you need help with. (and why?)
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