Calculating pH and pOH

Now that we know what the pH and pOH scales are and how to calculate each value let’s look at how we determine the pH of a solution given an acid or a base concentration.

The Intuitive Approach

When a question asks us for the pH of a solution it is best to begin by determining which way we expect our solution to go. For example, if a solution starts at a pH of 7 and we add acid to the solution I expect its pH to go down and its pOH to go up. While this might seem obvious you can almost always eliminate answers by simply taking a step back and asking what are we adding?

It is a great first step because if you are lost on the remaining steps at least you have a better chance of guessing the correct answer. From here we will then ask did we add a strong or weak acid or base. Strong acids and bases tend to lead to larger changes in pH and pOH while weaker ones lead to smaller changes in both scales. Therefore we would expect a strong acid to change the pH far more than a weaker one.

This isn’t foolproof though since a really small amount of a strong acid might change the pH by a small amount while a large quantity of a weak acid might profoundly affect the pH. Again this is only going to guide our answer choices not decide them so long as we can figure out how to solve the problem more definitively.

Strong Acids and Bases

We will begin by learning this more definitive way of solving pH and pOH questions by starting with strong acids and bases since they are easier than weak ones. Since a strong acid/base is defined by its ability to completely dissociate into solution determining the H⁺ or OH^– concentration is a simple mole to mole ratio.

[Latexpage]
\[HCl\;(aq)\rightleftharpoons H^+\; (aq)\; + \; Cl^-\; (aq)\]
\[NaOH\;(aq)\rightleftharpoons Na^+\; (aq)\; + \; OH^-\; (aq)\]

For example a 10M HCl solution will go 100% to products and produce a 10M concentration of both H⁺ and Cl^– ions. The same is true for a 1M NaOH solution except here it will produce a 1M OH^– and 1M Na⁺ concentration instead. Under most circumstances we can simply plug this value into our pH or pOH equation and disregard the H⁺ or OH^– ions present in water. Here is an example using the 10M HCl example:

[Latexpage]
\[pH=-log[H^{+}]\]
\[pH=-log[10^{1}]\]
\[pH=-1\]

This works because the by comparison the amount of either ion produced by the acid or base is far greater than water’s concentration of either. Technically the H⁺ concentration is equal to the 10M from HCl and the 1×10^-7 M from water or 10.0000001M which is basically 10M so it doesn’t matter. Another important thing to note here is that the pH and pOH scale don’t end at 14 and 0. Negative pH and pOH values are perfectly acceptable but they indicate extremely acidic or basic solutions.

However when very small quantities of a strong acid or base are added to solution the acid base concentrations in water do matter. For instance if we added 1×10^-8 M NaOH to solution it would produce 1×10^-8 M OH^–. Here the concentration of additional OH^– atoms is less than waters so the new concentration of hydroxide ions in solution is 1.1×10^-7.

Let’s imagine for a moment we forgot this fact and calculated the pH of this solution using the concentration of NaOH alone. This would result in a pOH of 8 and a pH of 6.

[latexpage]
\[pOH=-log[OH^-]\]
\[pOH=-log[1\times10^{-8}]\]
\[pOH=8\]
\[pOH+pH=14\]
\[pH=14-8=6\]

We added base to the solution so a pH of 6 makes no sense. How can a base increase the amount of H⁺ ions in solution if it’s OH^– ions are actively neutralizing them. It can’t we just forgot to account for hydroxide concentration already present in water. If we use the correct amount (1.1×10^-7 M OH^–) we would find that the actual pOH of the solution is around 6.89 and therefore the new pH is actually 7.11.

[latexpage]
\[pOH = -log[OH^-]\]
\[pOH=-log [1.1\times^{-7}]\]
\[pOH = 7 – \frac{1.1}{10}=7-0.11=6.89\]

Thankfully the MCAT doesn’t expect us to be able to calculate logs with great accuracy so you should be able to recognize the correct answer using the trick we covered in the log topic a few lessons back rather than needing an exact answer.

Weak Acids and Bases

Weak acids and bases are a bit trickier and we can’t use their mole to mole ratios because they don’t dissociate completely. Instead they exist as a mixture of acid and conjugate base or base and conjugate acid at equilibrium. Therefore we will need to determine their H⁺ and OH^– concentrations using an equilibrium constant and the dreaded ICE table.

For an acid we will use a special K value called the acid dissociation constant (Ka).

[Latexpage]
\[ HA \rightleftharpoons H^+ + A^-\]
As always K values are equal to the products over the reactants and water is omitted from the equilibrium equation.
\[K=\frac{[Products]}{[Reactants]}\]
So the acid dissociation constant is equal to:
\[K_a=\frac{[H^+][A^-]}{[HA]}\]

Bases have a nearly identical equilibrium constant except they are dealing with the production of hydroxide ions instead of H⁺ ions. Their K value is called the base dissociation constant (Kb).

[Latexpage]
\[ BOH \rightleftharpoons B^+ + OH^-\]
Again a K value is equal to the products over the reactants.
\[K=\frac{[Products]}{[Reactants]}\]
So the base dissociation constant is equal to:
\[K_b=\frac{[B^+][OH^-]}{[BOH]}\]

We can also write out these expressions for an acid and base pair. To see how this works let’s consider the bicarbonate buffering reactions that our blood uses to regulate pH. Here we will begin by looking at the acidic form of the reaction

[Latexpage]
\[ HCO_3^-+ H_2O \rightleftharpoons H_3O^+ + CO_3^{2-}\]
As always a K value is equal to products over reactants.
\[K=\frac{[Products]}{[Reactants]}\]
So the acid dissociation constant for the bicarbonate buffer is:
\[K_a=\frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^- ]}\]

Now let’s look at the base form of the reaction:

[Latexpage]
\[ CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^- \]
As always a K value is equal to products over reactants.
\[K=\frac{[Products]}{[Reactants]}\]
So the base dissociation constant for the bicarbonate buffer is:
\[K_b=\frac{[OH^-][HCO_3^- ]}{[CO_3^{2-}]}\]

In either case, we should recognize that acids or bases that produce more [H+] or [OH-] will have higher K values. This makes sense. Strong acids and bases favor dissociation. As a result, they produce more hydrogen and hydroxide ions increasing the numerator of their respective K constants. Thus larger K_a values indicate stronger acids and larger K_b values stronger bases.

Scientists wanted to be able to compare K_a and K_b values more easily so they converted them pKa and pKb values. This puts them on the same scale as pH and pOH meaning that as the pKa value of an acid decreases its acidity increases and vice versa. The same is true for pKb except here as the pKb goes down the basicity of the compound increases and vice versa.

[latexpage]
\[pK_a = -log(K_a)\]
\[pK_b= -log(K_b)\]

Furthermore, if we were to combine any weak acid and weak base reaction we would find that all of the reactions can be distilled down to the auto-ionization of water.

This is important because it means we can relate K_b and K_a to one another in the same way as the [H⁺] and [OH^–]. So the K_a of an acid and the K_b of its conjugate base will always equal K_w or 1 x10^-14. This also means that the pK_b and the pK_a like pH and pOH must always equal 14.

Calculating pH and pOH for Weak Acids and Bases

Great, we know about Ka, Kb, pKa, and pKb values but how does this help us with pH and pOH calculations? It allows us to determine the equilibrium concentrations of hydroxide or the hydronium ion concentration which can then be used to find the pH or pOH of a solution.

To set up this equation we will need to create an ICE table. Although they can be a bit intimidating at first the key is to mastering the setup and recognizing that the MCAT will always use the simplification assumption. To get started begin by writing out the balanced chemical equation above the ICE mnemonic.

From here define the equilibrium constant in terms of the concentrations of the products and reactants. Remember K values are always products over reactants and water is excluded from the expression.

Now begin filling in the initial concentration of any of the components in the equation. If you are only given the concentration of the reactants then the product’s initial concentration is zero. From there write in the change in each component. We will always skip the reactant’s change because we aren’t expected to carry out quadratics on the exam. For the products write the change in concentration as +x since we will be solving for this value.

Now carry down the values from both the I and C steps to E.

Lastly, fill in the Ka expression with the matching values from the E step of the ICE table.

Now we can solve for X using a given K_a value. To see how this works let’s use the K_a value of 1.8 x10^-5 for acetic acid.

While the math might seem intimidating we can easily break this problem down if we focus on two key rules. First, always rearrange your decimal points so you have an even exponent. Second, look for perfect squares within the numbers you are given.