Question

Formic acid (HCOOH) has Ka= 1.8x10-4. What is the pH of a 0.01 M solution of formic acid ?

PLEASE HELP I HAVE EXAM

Answer 1

Formic acid (CH2O2) partially dissociates in water according to the equation:
CH2O2 (aq) ? H+ (aq) + CH2O- (aq)
Ka is therefore:
Ka = [CH2O-] [H+] / [CH2O2]
The ICE table is:
............... ........[CH2O2]........ ........[CH2O-]...... .....[H+]
Initial......... .......0.01 M.......... ............0 M....... .......0
Change...... ........-x........... ................+x......... ........+x
Equilibrium.. ....0.01 M - x..... ..............x......... .........x
Substitute the E line into the Ka equation:
1.8x10^-4 = x *x / (0.01 M - x)
Since Ka is small in this case, so is x, and we can ignore x in the denominator to give:
1.8x10^-4 = x^2 / 0.01 M
Solving, [H+] = 1.341x10^-3 M, so pH = 2.8725
So in general, for a weak acid of known Ka,
[H+] = ?[Ka*c],
where c is the concentration of the weak acid. Formic acid is just about the strongest acid where we can get away with this shortcut, any stronger (larger Ka) and it's more accurate to rearrange the initial E line substitution into a quadratic.