Thursday
March 23, 2017

Post a New Question

Posted by 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 - , 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?)

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question