Question

at 25 C, a 0.010mol/L ammonia solution is 4.3 percent ionized. Calculate the pH

Answer 1

I like to focus on the reaction and what is happening - ammonia (NH3) is a weak base so it will only partially ionize as shown below

NH3(aq) + H2O(l) <----> NH4+ (aq) + OH-(aq)

So, if we start with 0.01 M NH3, what does it exactly mean when it is 4.3% ionized? This means that out of the original 0.01 M sample of NH3, only 4.3% is converted to NH4+

Amount of NH3 converted to NH4+: (4.3/100)(0.01) = 0.00043 M

So if we construct the following chart for the original reaction

NH3(aq) + H2O(l) <----> NH4+ (aq) + OH-(aq)

Initial molarity 0.01 mole/L 0 mole/L 0 mole/L

change in molarity - 0.00043 mole/L +0.00043 M 0.00043 mole/L

molarity at equilibrium 0.00957 M 0.00043 M 0.00043 M

Note: the change in the molarity for OH- is the same as the change in molarity for NH4+ because they have the same reaction coefficients

pH really depends on the H+ molarity

This is a useful equation: 1x10-14 = [H+][OH-] = [H+][0.00043 M]

[H+] = 1x10-14 / 0.00043 = 2.3x10-11

pH = -log[H+] = -log(2.3x10-11)