pH

What is pH?

In Brønsted-Lowry Acids and Bases, we defined an acid as a proton donor. Protons are just hydrogen ions, ${\mathrm{H}}^{+}$. The stronger an acid is, the better it is at donating protons, the lower its pH. Using the scale above, we can see that balsamic vinegar is a much stronger acid than soap, for example – it donates more protons in solution.

We can represent pH with the following equation:

$\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right(\mathrm{aq}\left)\right]$

So if we know the concentration of hydrogen ions in a solution, we can calculate the pH of the solution.

The higher the concentration of hydrogen ions in a solution, the lower the pH, and vice versa. A pH of less than 7 is acidic, whilst more than 7 is alkaline. You may have been told in the past that a pH of 7 is neutral, but in reality, neutral has a different definition.

What are strong acids and bases?

You may recall from the previous article that acids are proton donors. In solution, they dissociate to form negative ions and positive hydrogen ions. This dissociation is also called ionisation. It is a reversible reaction, as shown below:

$\mathrm{HX}\leftrightharpoons {\mathrm{H}}^{+}+{\mathrm{X}}^{-}$

However, some acids are very good at giving up their hydrogen ions -– so good that the reaction is essentially one-way. We call these acids strong acids.

The following equation arises:

$\mathrm{HX}\to {\mathrm{H}}^{+}+{\mathrm{X}}^{-}$

Similarly, we can obtain strong bases.

If we add a strong base to water, we get the following equation:

$\mathrm{B}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{BH}}^{+}+{\mathrm{OH}}^{-}$

The pH value of strong acids and bases

Strong acids have a low pH because they have a high concentration of hydrogen ions in the solution. Examples are hydrochloric acid ($\mathrm{HCl}$), nitric acid (${\mathrm{HNO}}_3$) and sulfuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$).

Strong bases have a low concentration of hydrogen ions in the solution. As a result, they have a high pH. Examples of this are all group 1 and group 2 hydroxides, such as sodium hydroxide ($\mathrm{NaOH}$).

The typical pH ranges of strong acids and bases. Anna Brewer, StudySmarter Originals

How do we find the pH of strong acids?

Remember that a strong acid dissociates completely in an aqueous solution. Let us take hydrochloric acid as an example.

If we put $0.1$ moles of hydrochloric acid in $0.5{\mathrm{dm}}^3$ of water, the acid will completely dissociate into $0.1$ mole of hydrogen ions, ${\mathrm{H}}^{+}$, and 0.1 moles of chloride ions, ${\mathrm{Cl}}^{-}$.

To find the concentration, we divide the number of moles by the volume of the solution. So, to find the hydrogen ions concentration in this particular solution, we do $0.1\mathrm{፥}0.5=0.2\mathrm{mol}{\mathrm{dm}}^{-3}$.

Now what? How can we find the pH value?

Well, let's go back to our original equation for pH:

$\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right(\mathrm{aq}\left)\right]$

We now know the concentration of hydrogen ions in the solution. Therefore, we can substitute this into our equation as shown:

$\mathrm{pH}=-{\log }_{10}(0.2)=0.70$

Finding [H+] from pH

What if you know the pH of a strong acid and want to find its hydrogen ion concentration? We can rearrange the equation for pH to make $\left[{\mathrm{H}}^{+}\right]$ the subject. First, switch the minus sign around:

$-\mathrm{pH}={\log }_{10}\left(\right[{\mathrm{H}}^{+}\left]\right)$

Next, take antilogs of both sides:

${10}^{-\mathrm{pH}}=\left[{\mathrm{H}}^{+}\right]$

There you go! To find the hydrogen ion concentration, simply substitute your value for pH into the equation. Here's an example to help you understand more clearly:

Now, we know the pH. Let's substitute it into our equation:

$$\begin{gathered} {10}^{-0.75}=\left[{\mathrm{H}}^{+}\right] \\ \left[{\mathrm{H}}^{+}\right]=0.1778 \end{gathered}$$

This gives us the concentration of hydrogen ions in the solution. But, what do we know about hydrochloric acid? Hydrochloric acid is a strong acid and dissociates entirely in solution. Therefore, the concentration of hydrochloric acid is also $0.1778$. To find the number of moles of hydrochloric acid dissolved in the solution, we can multiply the concentration by the volume:

$0.1778\mathrm{x}0.5=0.089\mathrm{mols}$ to two decimal places.

How do we find the pH of strong bases?

Calculating the pH of strong bases is a little more tricky than working out the pH of a strong acid – there is an extra step. To do this, you need a value known as ${\mathrm{K}}_{\mathrm{w}}$. ${\mathrm{K}}_{\mathrm{w}}$ is also known as the Ionic Product of Water. It has the following equation:

${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

${\mathrm{K}}_{\mathrm{w}}$ varies depending on temperature. At a fixed temperature, ${\mathrm{K}}_{\mathrm{w}}$ always remains the same. For example, you'll usually work with acids and bases at room temperature – about 25℃. At this temperature, ${\mathrm{K}}_{\mathrm{w}}$ takes the value $1.00\mathrm{x}{10}^{-14}{\mathrm{mol}}^2{\mathrm{dm}}^{-6}$ .

If we know the concentration of hydroxide ions in the solution, we can use ${\mathrm{K}}_{\mathrm{w}}$ to find the concentration of hydrogen ions. We can then calculate the pH of the solution, as we did above.

Example: Calculate the pH of a $0.1\mathrm{mol}{\mathrm{dm}}^{-3}$ solution of sodium hydroxide, $\mathrm{NaOH}$.

Have a go at answering the question by yourself first. But if you are stuck, let us go through it together now.

Since sodium hydroxide is a strong base, it completely dissociates in solution into hydroxide ions and sodium ions, ${\mathrm{OH}}^{-}$ and ${\mathrm{Na}}^{+}$ respectively, as we explored above. The concentration of hydroxide ions in the solution is thus also $0.1\mathrm{mol}{\mathrm{dm}}^{-3}$. We can use this value, alongside ${\mathrm{K}}_{\mathrm{w}}$, to find the concentration of hydrogen ions in the solution.

${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

Divide both sides by $\left[{\mathrm{OH}}^{-}\right]$:

$\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]}=\left[{\mathrm{H}}^{+}\right]$

Substitute our values in:

$\frac{1.00\mathrm{x}{10}^{-14}}{0.1}=\left[{\mathrm{H}}^{+}\right]=1.00\mathrm{x}{10}^{-13}$

We can then put this into our equation for pH:

$$\begin{gathered} \mathrm{pH}=-\log (1.00\mathrm{x}{10}^{-13}) \\ \mathrm{pH}=13.00 \end{gathered}$$

How do we find the pH of mixtures?

Calculating the pH of an acid or base is all well and good, but in chemistry and everyday life, you'll more often encounter mixtures of acids and bases. These react with each other in neutralisation reactions. For example, if you suffer from heartburn, you might take magnesium hydroxide tablets to neutralise excess stomach acid. In this example, some of the hydroxide ions from the base, magnesium hydroxide, react with hydrogen ions from the acid, hydrochloric acid, to form water. They'll continue to react until one of the reagents is used up. This reagent is the limiting reactor, whilst the other reagent is in excess.

To determine the pH of a mixture, you need to determine which reagent is in excess. You can then calculate the concentration of the hydrogen ions or hydroxide ions remaining and then calculate the pH as before. We'll now go through an example to help you understand the process.

Example: A mixture contains $50{\mathrm{cm}}^3$ of $0.100\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{H}}_2{\mathrm{SO}}_4$ and $25{\mathrm{cm}}^3$ of $0.150\mathrm{mol}{\mathrm{dm}}^{-3}\mathrm{NaOH}$. Calculate its pH.

First of all, we need to convert volume into ${\mathrm{dm}}^3$:

$$\begin{gathered} 50{\mathrm{cm}}^3{\mathrm{H}}_2{\mathrm{SO}}_4=0.05{\mathrm{dm}}^3 \\ 0.05\mathrm{x}0.100=5\mathrm{x}{10}^{-3}\mathrm{moles}{\mathrm{H}}_2{\mathrm{SO}}_4 \end{gathered}$$

Then we can calculate the number of moles of hydrogen ions and hydroxide ions in the solution. However, you'll notice that H2SO4 contains two hydrogen atoms per mole, meaning each mole dissolved in the solution produces two moles of aqueous hydrogen ions. So we have $(5\mathrm{x}{10}^{-3})\mathrm{x}2=1\mathrm{x}{10}^{-2}$ moles of H+.

Let's do the same thing with $\mathrm{NaOH}$.

$$\begin{gathered} 25{\mathrm{cm}}^3\mathrm{NaOH}=0.025{\mathrm{dm}}^3 \\ 0.025\mathrm{x}0.150=3.75\mathrm{x}{10}^{-3}\mathrm{moles}\mathrm{NaOH}. \end{gathered}$$

Each mole of $\mathrm{NaOH}$ dissociates to produce just one mole of hydroxide ions, so we have $3.75\mathrm{x}{10}^{-3}$ moles of hydroxide ions.

But this is a neutralisation reaction, and so the hydrogen ions react with the hydroxide ions, as we mentioned above. They react in a 1:1 ratio. We have fewer hydroxide ions than hydrogen ions, which means that some hydrogen ions won't react and will remain in the solution. To find out this value, subtract the number of hydroxide ions from the number of hydrogen ions:

$(1\mathrm{x}{10}^{-2})-(3.75\mathrm{x}{10}^{-3})=6.25\mathrm{x}{10}^{-3}$

This is the number of hydrogen ions remaining in the solution. We can now calculate the pH as in the examples above, first by working out $\left[{\mathrm{H}}^{+}\right]$ using the total volume, which is $0.05+0.025=0.075{\mathrm{dm}}^3$, and then by taking logs:

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=(6.25\mathrm{x}{10}^{-3})\mathrm{፥}0.075{\mathrm{dm}}^3=0.083\mathrm{mol}{\mathrm{dm}}^{-3} \\ \mathrm{pH}=-\log (0.083)=1.08 \end{gathered}$$

The following flowchart summarises how you calculate pH for strong acids, bases, and mixtures of the two.

A flowchart for calculating pH. Anna Brewer, StudySmarter Originals