ABSE_pH of strong Acids and Bases Lesson

pH of strong Acids and Bases

At the beginning of this module, you learned how to use the pH equation. Did you notice that we only used this equation (and the pOH equation) on strong acids and bases? This is because more work must be done before this equation can be used to calculate the pH of weak acids and bases. Before we calculate pH of weak acids and bases, let's review calculating pH of strong acids and bases. In strong acids, we assume that acid molecule ionizes completely. So, the molarity of the acid gives us the molarity of the H+. This molarity of the H+ is what we use to calculate the pH.

When an acid is monoprotic, meaning that it has only one hydrogen available for ionization, the molarity of the solution is the same as the molarity of the H+.

Example: Find the pH of a 0.03 M solution of HCI. Since HCI is a monoprotic acid, the molarity of the solution is the same as the molarity of the H+. So, [H+] = [HCL] [H+] = 0.03 M pH = -log(0.03) pH = 1.5

When an acid is polyprotic, meaning that it has more than one hydrogen available for ionization, the molarity of the solution is NOT the same as the molarity of the H+. To determine the [H+], you will look at the hydrogen available for ionization. If this is not visible to you by inspection, you can write the dissociation equation.

Example: Find the pH of a 0.03 M solution of H₂SO₄. Since H2SO4 is polyprotic, the molarity of the solution is not the same as the molarity of the H+. You can see by inspection that the [H+] is 2 times that of [H₂SO₄] because there are 2 hydrogens available for ionization. If this is not visible by inspection to you, write the dissociation equation, like this: H₂SO4₂H⁺ + SO₄²⁻ So, [H+] = 0.03 x 2 = 0.06 M pH = -log(0.06) pH = 1.2

pH of Weak Acids and Bases

As we mentioned above, calculating the pH of weak acids and bases requires more steps than does calculating the pH of strong acids and bases. The reason for this is that we are usually given the molarity of the solution, not the molarity of the H+. [H+] was easy to determine in a strong acid because strong acids dissociate completely. Weak acids do not dissociate completely, so we must determine the amount of H+ that is actually present. To do this we need to determine the [H+] at equilibrium using an "ice" chart like we did in the previous module on equilibrium. Follow along with the example in this video to see how:

These Khan Academy videos work examples of calculating the pH of a weak acid and a weak base when the assumption mentioned above is made. Pay attention to the difference in the answer when the assumption is made versus when the quadratic equation is used.

In the first video, where the pH of HF is determined, how does the pH of the solution calculated with the assumption made compare to the pH of the solution calculated using the quadratic equation?

  • pH using the assumption is 1.49. The pH using the quadratic equation is also 1.49. The assumption is correct.

Sometimes you will be given ka when you need kb or vice versa. In these cases you either need to change the equation you write to match the equilibrium constant given or you need to calculate the equilibrium constant you need.   Here is an example:

Example An experiment used a 0.15 M NH3. Calculate the pH of the sample. K₁ = 5.56 x 10-10 We want to solve for pH. Since ammonia, NH3, is a weak base, we will need to use the equilibrium concentration of OH to calculate the pOH, then pH. pH = ? pH = -log[H*]eq We start by writing the ionization equation of this weak base. NH₂+ H2O NH4+*+ OH We were given ko, but we are using the ionization equation of the base. Therefore we will need to calculate ko Ka X K = Kw ka x 5.56 x 10-10 = 1.0 x 10-14 ko 1.8 x 10-5 k- [NH][OH]/[NH₃] = 1.8 x 10⁻⁵ (x)(x)/(0.15-x) = 1.8 x 10⁻⁵ The denominator can be assumed to be 0.15. (x)² (0.15) = 1.8 x 10 x = 1.6 x 10 = [OH⁻] pOH = -log[OH]eq pOH = -log (1.2 x 10-³) POH = 2.80 pH+ pOH = 14 pH = 11.2

Percent Ionization

Percent ionization is just a different way to describe how complete the reaction is by comparing the amount of ions formed to the amount of ions that could be formed.  

% IONIZATION % Ionization = moles ionized per liter/moles available per liter x 100

Percent ionization can be calculated from the equilibrium concentrations. For example, to calculate % ionization from the previous example, we use the [H+] at equilibrium and the original molarity (since the acid is monoprotic).  

LaTeX: \text{% ionization}=\frac{1.2x10^{-3}}{0.10}x100=1.2%% ionization=1.2x10−30.10x100=1.2

We can see that this is a very weak acid since only 1.2% of the molecules formed ions.

The pH of a weak acid or base can also be calculated from the percent ionization.  

Example A 0.150 M solution of acetic acid (shorthand formula = HAc) is found to be 1.086 % ionized. What is the pH? HC₂H₃O₂H+ + C₂H₃O2* (Acetate is often abbreviated Ac. This abbreviation is used below.) HAC H++ Ac We can use the % ionization to determine the [H+] at equilibrium. % Ionization = [H] eq moles available per liter x 100 1.086% [H⁺], eq 0.150 x 100 [H+] 1.63 x 10⁻³ This [H+] is used to calculate pH. pH = -log (1.63 x 10⁻³) pH = 2.79

Calculating ka and kb

If given the pH, pOH, or percent ionization of a weak acid or base, we can work backwards to calculate ka or kb.   Follow along with the example in the video.

Here is an example calculating ka given pH.

Example Formic Acid, HCHO2, has a concentration of 0.100 M, pH = 2.38. Calculate Ka. Write the ionization equation for HCHO₂ HCHO2 = H+ + CHO Set up an ice chart and write in the initial values. The equilibrium values can be plugged into the ka expression. [H][CHO ] [HCHO,] (4.2 x 10³) (4.2 x 10³) k,- = = 1.84 x 10" (0.096)

Remember to work on the module practice problems as you complete each section of content.  

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