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

For Further Reading

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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