# pH, pOH, and the pH scale

Definitions of pH, pOH, and the pH scale. Calculating the pH of a strong acid or base solution. The relationship between acid strength and the pH of a solution.

## Key points

• We can convert between $[\text{H}^+]$ and $\text{pH}$ using the following equations:
\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ [\text H^+]&=10^{-\text{pH}}\end{aligned}
• We can convert between $[\text{OH}^-]$ and $\text{pOH}$ using the following equations:
\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ [\text {OH}^-]&=10^{-\text{pOH}}\end{aligned}
• For any aqueous solution at $25\,^\circ\text{C}$:
$\text{pH}+\text{pOH}=14$.
• For every factor of $10$ increase in concentration of $[\text{H}^+]$, $\text{pH}$ will decrease by $1$ unit, and vice versa.
• Both acid strength and concentration determine $[\text{H}^+]$ and $\text{pH}$.

## Introduction

In aqueous solution, an acid is defined as any species that increases the concentration of $\text{H}^+(aq)$, while a base increases the concentration of $\text{OH}^-(aq)$. Typical concentrations of these ions in solution can be very small, and they also span a wide range.
Purple-blue hydrangeas next to pinkish purple hydrangeas.
The color of hydrangea flowers can vary depending on the pH of the soil. Blue flowers usually come from acidic soil with a pH less than 6, and and pink flowers come from soil with a pH above 6. Photo from WIkimedia Commons, CC BY 2.0
For example, a sample of pure water at $25\,^\circ\text{C}$ contains $1.0 \times 10^{-7}\text{ M}$ of $\text{H}^+$ and $\text{OH}^-$. In comparison, the concentration of ${\text H^+}$ in stomach acid can be up to ~$1.0 \times10^{-1}\,\text M$. That means $[\text{H}^+]$ for stomach acid is approximately $6$ orders of magnitude larger than in pure water!
To avoid dealing with such hairy numbers, scientists convert these concentrations to $\text{pH}$ or $\text{pOH}$ values. Let's look at the definitions of $\text{pH}$ and $\text{pOH}$.

## Definitions of $\text{pH}$ and $\text{pOH}$

### Relating $[\text H^+]$ and $\text{pH}$

The $\text{pH}$ for an aqueous solution is calculated from $[\text{H}^+]$ using the following equation:
$\text{pH}=-\log[\text{H}^+]\quad\quad\quad\quad\;\;\;\text{(Eq. 1a)}$
The lowercase $\text{p}$ indicates $-\text{log}_{10}"$. You will often see people leave off the base $10$ part as an abbreviation.
For example, if we have a solution with $[\text{H}^+]=1 \times 10^{-5}\text{ M}$, then we can calculate the $\text{pH}$ using Eq. 1a:
$\text{pH}=-\log(1 \times 10^{-5})=5.0$
Given the $\text{pH}$ of a solution, we can also find $[\text{H}^+]$:
$[\text H^+]=10^{-\text{pH}}\quad\quad\quad\quad\quad\;\;\;\text{(Eq. 1b)}$
There are a few rules you want to know for calculations involving logarithms like $\text {pH}$ and $\text {pOH}$. For a review of log properties, see Sal's video on properties of logarithms.

### Relating $[\text {OH}^-]$ and $\text{pOH}$

The $\text{pOH}$ for an aqueous solution is defined in the same way for $[\text{OH}^-]$:
$\text{pOH}=-\log[\text{OH}^-]~\quad\quad\quad\;\;\text{(Eq. 2a)}$
For example, if we have a solution with $[\text{OH}^-]=1 \times 10^{-12}\text{ M}$, then we can calculate $\text{pOH}$ using Eq. 2a:
$\text{pOH}=-\log(1 \times 10^{-12})=12.0$
Given the $\text{pOH}$ of a solution, we can also find $[\text{OH}^-]$:
$10^{-\text{pOH}}=[\text {OH}^-]\quad\quad\quad\quad\quad\;\;\;\text{(Eq. 2b)}$

### Relating $\text{pH}$ and $\text{pOH}$

Based on equilibrium concentrations of $\text{H}^+$ and $\text{OH}^-$ in water, the following relationship is true for any aqueous solution at $25\,^\circ\text{C}$:
$\text{pH}+\text{pOH}=14~~\quad\quad\quad\quad\quad\quad\quad\quad\text{(Eq. 3)}$
This relationship can be used to convert between $\text{pH}$ and $\text{pOH}$. In combination with Eq. 1a/b and Eq. 2a/b, we can always relate $\text{pOH}$ and/or $\text{pH}$ to $[\text{OH}^-]$ and $[\text{H}^+]$. For a derivation of this equation, check out the article on the autoionization of water.

## Example 1: Calculating the $\text{pH}$ of a strong base solution

If we use $1.0\text{ mmol}$ of $\text{NaOH}$ to make $1.0\text{ L}$ of an aqueous solution at $25\,^\circ\text{C}$, what is the $\text{pH}$ of this solution?
We can find the $\text{pH}$ of our $\text{NaOH}$ solution by using the relationship between $[\text{OH}^-]$, $\text{pH}$, and $\text{pOH}$. Let's go through the calculation step-by-step.

#### Step 1. Calculate the molar concentration of $\text{NaOH}$

Molar concentration is equal to moles of solute per liter of solution:
$\text{Molar concentration}=\dfrac{\text{mol solute}}{\text{L solution}}$
To calculate the molar concentration of $\text{NaOH}$, we can use the known values for the moles of $\text{NaOH}$ and the volume of solution:
\begin{aligned}\text{[NaOH]}&=\dfrac{1.0\text{ mmol NaOH}}{1.0\text{ L}}\\ \\ &=\dfrac{1.0\times10^{-3}\text{ mol NaOH}}{1.0\text{ L}}\\ \\ &=1.0\times10^{-3}\text{ M NaOH}\end{aligned}
The concentration of $\text{NaOH}$ in the solution is $1.0\times10^{-3}\text{ M}$.

#### Step 2: Calculate $[\text{OH}^-]$ based on the dissociation of $\text{NaOH}$

Because $\text{NaOH}$ is a strong base, it dissociates completely into its constituent ions in aqueous solution:
$\text{NaOH}(aq)\rightarrow\text{Na}^+(aq)+\text{OH}^-(aq)$
This balanced equation tells us that every mole of $\text{NaOH}$ produces one mole of $\text{OH}^-$ in aqueous solution. Therefore, we have the following relationship between $[\text{NaOH}]$ and $[\text{OH}^-]$:
$[\text{NaOH}]=[\text{OH}^-]=1.0\times10^{-3}\text{ M}$

#### Step 3: Calculate $\text{pOH}$ from $[\text{OH}^-]$ using Eq. 2a

Now that we know the concentration of $\text{OH}^-$, we can calculate $\text{pOH}$ using Eq. 2a:
\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ &=-\log(1.0\times10^{-3})\\ \\ &=3.00\end{aligned}
The $\text{pOH}$ of our solution is $3.00$.

#### Step 4: Calculate $\text{pH}$ from $\text{pOH}$ using Eq. 3

We can calculate $\text{pH}$ from $\text{pOH}$ using Eq. 3. Rearranging to solve for our unknown, $\text{pH}$:
$\text{pH}=14-\text{pOH}$
We can substitute the value of $\text{pOH}$ we found in Step 3 to find the $\text{pH}$:
$\text{pH}=14-3.00=11.00$
Therefore, the $\text{pH}$ of our $\text{NaOH}$ solution is $11.00$.

## The $\text{pH}$ scale: Acidic, basic, and neutral solutions

Converting $[\text{H}^+]$ to $\text{pH}$ is a convenient way to gauge the relative acidity or basicity of a solution. The $\text{pH}$ scale allows us to easily rank different substances by their $\text{pH}$ value.
The $\text{pH}$ scale is a negative logarithmic scale. The logarithmic part means that $\text{pH}$ changes by $1$ unit for every factor of $10$ change in concentration of $\text{H}^+$. The negative sign in front of the log tells us that there is an inverse relationship between $\text{pH}$ and $[\text{H}^+]$: when $\text{pH}$ increases, $[\text{H}^+]$ decreases, and vice versa.
The following image shows a $\text{pH}$ scale labeled with $\text{pH}$ values for some common household substances. These $\text{pH}$ values are for solutions at $25\,^\circ\text{C}$. Note that it is possible to have a negative $\text{pH}$ value.
The pH scale from -1 to 14.
The pH scale. Acidic solutions have pH values less than 7, and basic solutions have pH values greater than 7. Image from UCDavis ChemWiki, CC BY-NC-SA 3.0 US.
Some important terminology to remember for aqueous solutions at $25\,^\circ\text{C}$:
• For a neutral solution, $\text{pH}=7$.
• Acidic solutions have $\text{pH}<7$.
• Basic solutions have $\text{pH}>7$.
The lower the $\text{pH}$ value, the more acidic the solution and the higher the concentration of $\text H^+$. The higher the $\text{pH}$ value, the more basic the solution and the lower the concentration of $\text H^+$. While we could also describe the acidity or basicity of a solution in terms of $\text{pOH}$, it is a little more common to use $\text{pH}$. Luckily, we can easily convert between $\text{pH}$ and $\text{pOH}$ values.
Concept check: Based on the $\text{pH}$ scale given above, which solution is more acidic$-$orange juice, or vinegar?
Vinegar has a $\text{pH}$ of $\approx3.0$, and orange juice has a $\text{pH}$ of $\approx3.6$. Because the $\text{pH}$ of vinegar is less than that of orange juice $(3.0<3.6)$, vinegar has a higher concentration of $\text{H}^+$.
Therefore, vinegar is more acidic than orange juice.

## Example $2$: Determining the $\text{pH}$ of a diluted strong acid solution

We have $100\text{ mL}$ of a nitric acid solution with a $\text{pH}$ of $4.0$. We dilute the solution by adding water to get a total volume of $1.0\text{ L}$.
What is the $\text{pH}$ of the diluted solution?
There are multiple ways to solve this problem. We will go over two different methods.

### Method 1. Use properties of the log scale

Recall that $\text{pH}$ scale is a negative logarithmic scale. Therefore, if the concentration of $\text{H}^+$ decreases by a single factor of $10$, then the $\text{pH}$ will increase by $1$ unit.
Since the original volume, $100\text{ mL}$, is one tenth the total volume after dilution, the concentration of $\text H^+$ in solution has been reduced by a factor of $10$. Therefore, the $\text{pH}$ of the solution will increase $1$ unit:
\begin{aligned}\text{pH}&=\text{original pH}+1.0 \\ \\ &=4.0+1.0\\ \\ &=5.0\end{aligned}
Therefore, the $\text{pH}$ of the diluted solution is $5.0$.

### Method 2. Use moles of $\text{H}^+$ to calculate $\text{pH}$

#### Step 1: Calculate moles of $\text{H}^+$

We can use the $\text{pH}$ and volume of the original solution to calculate the moles of $\text{H}^+$ in the solution.
\begin{aligned}\text{moles H}^+&=[\text H^+]_{initial} \times \text{volume}\\ \\ &=10^{-\text{pH}}\,\text M \times \text {volume}\\ \\ &=10^{-4.0}\,\text M \times {0.100\text{ L}}\\ \\ &=1.0 \times 10^{-5}\,\text {mol H}^+\end{aligned}

#### Step 2: Calculate molarity of $\text{H}^+$ after dilution

The molarity of the diluted solution can be calculated by using the moles of $\text{H}^+$ from the original solution and the total volume after dilution.
\begin{aligned}[\text{H}^+]_{final}&=\dfrac{\text{mol H}^+}{\text{L solution}}\\ \\ &=\dfrac{1.0 \times 10^{-5}\,\text {mol H}^+}{1.0\,\text{L}}\\ \\ &=1.0 \times 10^{-5}\,\text M\end{aligned}

#### Step 3: Calculate $\text{pH}$ from $[\text{H}^+]$

Finally, we can use Eq. 1a to calculate $\text{pH}$:
\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ &=-\text{log}(1.0 \times 10^{-5})\\ \\ &=5.0\end{aligned}
Method 2 gives us the same answer as Method 1, hooray!
In general, Method 2 takes a few extra steps, but it can always be used to find changes in $\text{pH}$. Method 1 is a handy shortcut when changes in concentration occur as multiples of $10$. Method 1 can also be used as a quick way to estimate $\text{pH}$ changes.

## Relationship between $\text{pH}$ and acid strength

Based on the equation for $\text{pH}$, we know that $\text{pH}$ is related to $[\text{H}^+]$. However, it is important to remember that $\text{pH}$ is not always directly related to acid strength.
The strength of an acid depends on the amount that the acid dissociates in solution: the stronger the acid, the higher $[\text{H}^+]$ at a given acid concentration. For example, a $1.0\,\text M$ solution of strong acid $\text{HCl}$ will have a higher concentration of $\text{H}^+$ than a $1.0\,\text M$ solution of weak acid $\text{HF}$. Thus, for two solutions of monoprotic acid at the same concentration, $\text{pH}$ will be proportional to acid strength.
More generally though, both acid strength and concentration determine $[\text{H}^+]$. Therefore, we can't always assume that the $\text{pH}$ of a strong acid solution will be lower than the $\text{pH}$ of a weak acid solution. The acid concentration matters too!

## Summary

Hand holding wet pH paper with four stripes (from left to right): orange, green-brown, yellow, and red-orange. The paper is held up for comparison against a reference chart of pH values and colors. The wet paper matches the pH 7 on the reference.
Indicator paper can be used to measure the pH of aqueous solutions. The color of the indicator paper in this picture matches a pH value of 7. Photo from Wikimedia Commons, CC BY-SA 2.5
• We can convert between $[\text{H}^+]$ and $\text{pH}$ using the following equations:
\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ [\text H^+]&=10^{-\text{pH}}\end{aligned}
• We can convert between $[\text{OH}^-]$ and $\text{pOH}$ using the following equations:
\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ [\text {OH}^-]&=10^{-\text{pOH}}\end{aligned}
• For every factor of $10$ increase in concentration of $[\text{H}^+]$, $\text{pH}$ will decrease by $1$ unit, and vice versa.
• For any aqueous solution at $25\,^\circ\text{C}$:
$\text{pH}+\text{pOH}=14$.
• Both acid strength and concentration determine $[\text{H}^+]$ and $\text{pH}$.

1. The pH Scale” from UC Davis ChemWiki, CC BY-NC-SA 3.0

Zumdahl, S.S., and Zumdahl S.A. (2003). Atomic Structure and Periodicity. In Chemistry (6th ed., pp. 290-94), Boston, MA: Houghton Mifflin Company.

## Problem $1$: Calculating the $\text{pH}$ of a strong base solution at $25\,^\circ\text{C}$

We make $200\text{ mL}$ of a solution with a $0.025\text{ M}$ concentration of $\text{Ca(OH)}_2$. The solution is then diluted to $1.00\text{ L}$ by adding additional water.
What is the $\text{pH}$ of the solution after dilution?
$\text{Ca(OH)}_2(aq)\rightarrow\text{Ca}^{2+}(aq)+2\text{OH}^-(aq)$
We see that $2\text{ mol OH}^-$ are produced for every $1\text{ mol Ca(OH)}_2$ that dissociates. Therefore, the concentration of hydroxide ion, $[\text{OH}^-]$, will be twice the molarity of $\text{Ca(OH)}_2$:
$[\text{OH}^-]_\text{initial}=0.025\text{ M}\times \dfrac{2\,\text{mol OH}^-}{1\,\text{mol Ca(OH)}_2}=0.050\text{ M}$
The solution has been diluted by a factor of $5$, since $200\text{ mL}\times5=1.0\text{ L}$. Therefore, the final molar concentration of $\text{OH}^-$ will be $\dfrac{1}{5}$ its initial value:
$[\text{OH}^-]_\text{final}=0.050\text{ M}\times\dfrac{1}{5}=0.010\text{ M}$
We can calculate the $\text{pOH}$ of the diluted solution from $[\text{OH}^-]$:
$\text{pOH}=-\log[\text{OH}^-]=-\log(0.010)=2.00$
Since $\text{pH}+\text{pOH}=14.00$ at $25\,^\circ\text{C}$,
\begin{aligned}\text{pH}&=14.00-\text{pOH}\\ \\ &=14.00-2.00\\ \\ &=12.00\end{aligned}
Therefore, the $\text{pH}$ of the diluted solution is $12.00$.