Known:
Cyanaocetic acid
Ka=3.55*10^-3
ionization equation: HC3H2NO2<==> H^+ + C3H2NO2^-
Unknown: pH of 0.4M of Cyanaocetic acid
How can I find out the pH of 0.4M of Cyanaocetic acid ? Is there another way solving for it other than using ICE chart? Please help me out.I really need this. Thanks in advance.
Let [H+] = x
Ka = [H^+][C3H2NO2^-] / [HC3H2NO2]
3.55x10^-3 = x^2 / (0.4-x)
•For an exact solution, you must convert the previous equation to a quadratic trinomial equal to 0, the solve.
•For an approximate solution, you assume that
(0.4-x)=0.4, when x is much smaller than 0.4.
Using the approximate method, 3.55x10^-3 = x^2 / 0.4.
x = [H+] = sqrt[(0.4)(3.55x10^-3)] = 0.0377 M
pH = 1.4
For a more precise answer, solve for x:
x^2 / (0.4-x) = 3.55x10^-3