Question

Calculate the pH (to two decimal places) of a 4.31×10^-3 M solution of oxalic acid. A table of pKa values can be found here. If the 5% approximation is valid, use the assumption to compute pH.

Answer 1

Solution:

Oxalic acid H₂C₂O₄ of concentraion 4.31 * 10^-3 M is given

Pka for oxaic acid you will get from standard values of Ka,

I have got the value of Ka = 5.9 * 10^-2

Use the ICE table to find out the equilibrium concentration of all species/

H₂C₂O₄ HC₂O₄^- + H⁺

Initial 4.31 * 10^-3 0 0

Change -x x x

Equilibri 4.31 * 10^-3 - x x x

Write down the expression for equilibrium constant;

Ka = [HC₂O₄^-][H⁺]/[H₂C₂O₄]

= x * x/(4.31 * 10^-3 - x)

Assume x is negligible

So,

5.9 * 10^-2 = x2/4.31 * 10^-3

x = 1.6 * 10^-2 M

As X is greater than 4.31 * 10^-3 M our assumption is wrong so we will recalculate x by considering it in the denominator,

5.9 * 10^-2 * (4.31 * 10^-3 - x) = x²

2.542 * 10^-4 - 5.9 * 10^-2x - x² = 0

x² + 5.9 * 10^-2x - 2.542 * 10^-4 = 0

Solve the quadratic euationfor x

x = 0.004032

[H⁺] = 0.004032

Therefore,

p^H = -log[H⁺]

= -log[0.004032]

p^H = 2.394