The general strategy for finding pH of strong acids and bases is:
1. Write the balanced reaction. Since a strong acid or base is involved, the reaction will always go to completion.
2. If the concentration of the acid or base is greater than 10–6 M, use the stoichiometry to find the H3O+ or OH– concentration.
3. Find the pH (acid) or pOH (base) using the definition (pH = –log[H3O+] or pOH = –log[OH–]).
4. For bases, to find the pH (at 25 °C) use pH = 14.00 – pOH.
Find the pH of a 0.0050 M HBr solution at 25 °C.
HBr is a strong acid, so:
HBr(aq) + H2O(l) → H3O+(aq) + Br–(aq)
Since the concentration of the strong acid exceeds 1.0×10–6 M, we can ignore the contribution from the water autoionization equilibrium. Thus,
[H3O+]e = 5.0×10–3M
pH = –log[H3O+] = –log(5.0×10–3) = 2.30
7.32×10–5 moles of NaOH is added to 2.0 L of water. What is the pH at 25 °C?
This is a strong base, so:
NaOH(aq) → Na+(aq) + OH–(aq)
[OH–] = 7.32×10–5 mol/2.0 L = 3.7×10–5 M
Since the concentration of the strong base exceeds 1.0×10–6 M, we can ignore the contribution from the water autoionization equilibrium. Thus,
pOH = –log(3.7×10–5) = 4.43
pH = 14.00 – 4.43 = 9.57
In general, we can write the reaction of a weak acid HA in aqueous as:
HA(aq) + H2O(l) → ← H3O+(aq) + A–(aq)
The equilibrium constant for this reaction is
Ka is the acid dissociation equilibrium constant.
Ka is a quantitative measure of acid strength:
A larger Ka means a stronger acid. (Why?)
Ka is often measured by one of two methods:
1. Measurement of the pH of a solution of known initial concentration of weak acid.
The pH of a solution gives the hydronium ion concentration at equilibrium. Using this information, the initial concentration, and the stoichiometry of reaction allows evaluation of all of the required equilibrium concentrations.
2. Measurement of the per cent ionization of a solution of known initial concentration of weak acid.
% ionization, given the notation α, is defined as
Again, this information and stoichiometry allows calculation of all equilibrium concentrations so that Ka can be evaluated.
Each of these methods is direct application of our typical methodology for treating equilibrium problems.
0.100 mole of HF is dissolved in 1.00 L of water at 25 °C. The pH at equilibrium was found to be 2.08. Evaluate Ka.
A 0.0100 M solution of HNO2 is found to be 19% ionized at equilibrium. Find Ka.
Which is the stronger acid, HF or HNO2?
The values of Ka range from 10–1 to 10–13 so a scale similar to pH is often used to report equilibrium constants:
pKa = –log Ka
So for HF and HNO2, the pKa values are:
HF pKa = –log Ka = –log(7.5×10–4) = 3.12
HNO2 pKa = –log Ka = –log(4.6×10–4) = 3.34
Smaller pKa means a stronger acid.
Acid ionization constants are reported in tables either as Ka or as pKa.
A table of Ka can be found here.
If an equilibrium constant is known, then the equilibrium concentrations of products and reactants can be determined. Usually, for weak acids, this means finding the pH of the solution.
This is a standard equilibrium problem, so we follow the typical procedures.
Find the pH of a 0.010 M solution of hypochlorous acid at 25 °C.
What is a reasonable answer to expect?
If this were a strong acid, the pH would be –log (0.010) = 2.00. If this were pure water, the pH would be 7.00. Thus, our answer must lie between these values.
What is the pH of a 3.8×10–5 M solution of acetic acid at 25 °C. pKa = 4.77
What is a reasonable answer to expect?
If this were a strong acid, the pH would be –log (3.8×10–5) = 4.42. If this were pure water, the pH would be 7.00. Thus, our answer must lie between these values. Further, since Ka is fairly large, the final answer probably is a little closer to 4 than to 7, perhaps ~5.
We can decide this by comparing Ka to the initial concentration:
If [HA]init Ka > 100
then approximation will give less than 5% error, which is acceptable.
If [HA]init Ka < 100
Then approximation is not acceptable and the calculation must be done using the quadratic formula.
