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# 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 $\left[{\text{H}}^{+}\right]$ and $\text{pH}$ using the following equations:
$\begin{array}{rl}\text{pH}& =-\mathrm{log}\left[{\text{H}}^{+}\right]\\ \\ \left[{\text{H}}^{+}\right]& ={10}^{-\text{pH}}\end{array}$
• We can convert between $\left[{\text{OH}}^{-}\right]$ and $\text{pOH}$ using the following equations:
$\begin{array}{rl}\text{pOH}& =-\mathrm{log}\left[{\text{OH}}^{-}\right]\\ \\ \left[{\text{OH}}^{-}\right]& ={10}^{-\text{pOH}}\end{array}$
• For any aqueous solution at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$:
$\text{pH}+\text{pOH}=14$.
• For every factor of $10$ increase in concentration of $\left[{\text{H}}^{+}\right]$, $\text{pH}$ will decrease by $1$ unit, and vice versa.
• Both acid strength and concentration determine $\left[{\text{H}}^{+}\right]$ and $\text{pH}$.

## Introduction

In aqueous solution, an acid is defined as any species that increases the concentration of ${\text{H}}^{+}\left(aq\right)$, while a base increases the concentration of ${\text{OH}}^{-}\left(aq\right)$. Typical concentrations of these ions in solution can be very small, and they also span a wide range.
For example, a sample of pure water at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$ contains of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$. In comparison, the concentration of ${\text{H}}^{+}$ in stomach acid can be up to approximately $1.0×{10}^{-1}\phantom{\rule{0.167em}{0ex}}\text{M}$. That means $\left[{\text{H}}^{+}\right]$ 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 $\left[{\text{H}}^{+}\right]$‍  and $\text{pH}$‍

The $\text{pH}$ for an aqueous solution is calculated from $\left[{\text{H}}^{+}\right]$ using the following equation:
$\text{pH}=-\mathrm{log}\left[{\text{H}}^{+}\right]\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\text{(Eq. 1a)}$
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 , then we can calculate the $\text{pH}$ using Eq. 1a:
$\text{pH}=-\mathrm{log}\left(1×{10}^{-5}\right)=5.0$
Given the $\text{pH}$ of a solution, we can also find $\left[{\text{H}}^{+}\right]$:
$\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\text{(Eq. 1b)}$

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

The $\text{pOH}$ for an aqueous solution is defined in the same way for $\left[{\text{OH}}^{-}\right]$:
For example, if we have a solution with , then we can calculate $\text{pOH}$ using Eq. 2a:
$\text{pOH}=-\mathrm{log}\left(1×{10}^{-12}\right)=12.0$
Given the $\text{pOH}$ of a solution, we can also find $\left[{\text{OH}}^{-}\right]$:
${10}^{-\text{pOH}}=\left[{\text{OH}}^{-}\right]\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\text{(Eq. 2b)}$

### 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 $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$:
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 $\left[{\text{OH}}^{-}\right]$ and $\left[{\text{H}}^{+}\right]$. 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 of $\text{NaOH}$ to make of an aqueous solution at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$, what is the $\text{pH}$ of this solution?
We can find the $\text{pH}$ of our $\text{NaOH}$ solution by using the relationship between $\left[{\text{OH}}^{-}\right]$, $\text{pH}$, 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:
$\text{Molar concentration}=\frac{\text{mol solute}}{\text{L solution}}$
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:
The concentration of $\text{NaOH}$ in the solution is .

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

Because $\text{NaOH}$ is a strong base, it dissociates completely into its constituent ions in aqueous solution:
$\text{NaOH}\left(aq\right)\to {\text{Na}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$
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 $\left[\text{NaOH}\right]$ and $\left[{\text{OH}}^{-}\right]$:

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

Now that we know the concentration of ${\text{OH}}^{-}$, we can calculate $\text{pOH}$ using Eq. 2a:
$\begin{array}{rl}\text{pOH}& =-\mathrm{log}\left[{\text{OH}}^{-}\right]\\ \\ & =-\mathrm{log}\left(1.0×{10}^{-3}\right)\\ \\ & =3.00\end{array}$
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 $\left[{\text{H}}^{+}\right]$ 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 $\left[{\text{H}}^{+}\right]$: when $\text{pH}$ increases, $\left[{\text{H}}^{+}\right]$ 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 $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$. Note that it is possible to have a negative $\text{pH}$ value.
Some important terminology to remember for aqueous solutions at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$:
• 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?

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

We have 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 .
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, , 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:
$\begin{array}{rl}\text{pH}& =\text{original pH}+1.0\\ \\ & =4.0+1.0\\ \\ & =5.0\end{array}$
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.

#### 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.
$\begin{array}{rl}\left[{\text{H}}^{+}{\right]}_{final}& =\frac{{\text{mol H}}^{+}}{\text{L solution}}\\ \\ & =\frac{1.0×{10}^{-5}\phantom{\rule{0.167em}{0ex}}{\text{mol H}}^{+}}{1.0\phantom{\rule{0.167em}{0ex}}\text{L}}\\ \\ & =1.0×{10}^{-5}\phantom{\rule{0.167em}{0ex}}\text{M}\end{array}$

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

Finally, we can use Eq. 1a to calculate $\text{pH}$:
$\begin{array}{rl}\text{pH}& =-\mathrm{log}\left[{\text{H}}^{+}\right]\\ \\ & =-\text{log}\left(1.0×{10}^{-5}\right)\\ \\ & =5.0\end{array}$
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 $\left[{\text{H}}^{+}\right]$. 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 $\left[{\text{H}}^{+}\right]$ at a given acid concentration. For example, a $1.0\phantom{\rule{0.167em}{0ex}}\text{M}$ solution of strong acid $\text{HCl}$ will have a higher concentration of ${\text{H}}^{+}$ than a $1.0\phantom{\rule{0.167em}{0ex}}\text{M}$ 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 $\left[{\text{H}}^{+}\right]$. 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

• We can convert between $\left[{\text{H}}^{+}\right]$ and $\text{pH}$ using the following equations:
$\begin{array}{rl}\text{pH}& =-\mathrm{log}\left[{\text{H}}^{+}\right]\\ \\ \left[{\text{H}}^{+}\right]& ={10}^{-\text{pH}}\end{array}$
• We can convert between $\left[{\text{OH}}^{-}\right]$ and $\text{pOH}$ using the following equations:
$\begin{array}{rl}\text{pOH}& =-\mathrm{log}\left[{\text{OH}}^{-}\right]\\ \\ \left[{\text{OH}}^{-}\right]& ={10}^{-\text{pOH}}\end{array}$
• For every factor of $10$ increase in concentration of $\left[{\text{H}}^{+}\right]$, $\text{pH}$ will decrease by $1$ unit, and vice versa.
• For any aqueous solution at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$:
$\text{pH}+\text{pOH}=14$.
• Both acid strength and concentration determine $\left[{\text{H}}^{+}\right]$ and $\text{pH}$.

## Problem 1: Calculating the pH of a strong base solution at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$‍

We make of a solution with a concentration of ${\text{Ca(OH)}}_{2}$. The solution is then diluted to by adding additional water.
What is the $\text{pH}$ of the solution after dilution?

## Want to join the conversation?

• How does the temperature affect the pH and pOH? •  At 100C the pH of water is 6.14, so higher temperature decreases the pH. The opposite is true for pOH: higher temperatures increases the pOH.
• What does M stand for in the unit labels? • Could someone explain the difference between acid strength and concentration? According to me, a strong acid will fully ionise in water compared to a weak acid which will partially ionise. Therefore a strong acid will contribute more H+ ions than a weak acid. Therefore, the pH of a strong acid solution will be higher than a weak acid solution.

Is this correct? •  Nice question!! It is important that you don't confuse the words strong and weak with the terms concentrated and dilute. At the same concentration, a weak acid will be less acidic than a strong acid. However, if you have highly concentrated weak acid (almost pure) and compare this to a very diluted strong acid (like 1 drop of HCl in a swimming pool) then the pH of the weak acid will be much more acidic than that of the strong acid.
• Can someone please explain what are monoprotic and diprotic acids? Thanks. • how can we solve pH,pOH numericals without using scientific calculator during our examination? • How can NaOH have a pH scale? How can a base add H+ ions to the solution? It adds OH- ions right? • NaOH does not have a pH, but an aqueous solution of NaOH does.
Water contains both H⁺ and OH⁻ ions.
Adding NaOH increases the concentration of OH⁻ ions and decreases the concentration of H⁺ ions.
But there are always some H⁺ ions present, so aqueous NaOH solutions have a pH, usually between 7 and 14.
• How does pH+pOH= 14? Where does the random number come in? • Hi
I feel like there's a step missing.I'm not sure why the pH as an exponent is negative & where the minus sign comes from.I understand that the logarithm (of base 10) was changed to an exponent.What is this law of logarithms called?
pH= -log (H+)
10^-pH = (H+)

Also I was trying to figure it out with numbers pH=-(log 10^-4)
and I got 10^-pH =10^-4 and I'm not sure where to go from there to obtain the pH.Do I just cancel out the 10s & minuses that are on both sides to get a pH of 4,to cross off these I have to divide/multiply both sides by some number(s) would these numbers be 10 and multiplying the exponents by -1 to get rid of the minuses because the pH scale is usually positive numbers?
Thanks! • This is the power rule of logs. When you have a number in front of a log term, this is the same as raising the log term to that number. For example, 4log(3) is the same as log(3^4).

With pH, the number in front of the log is -1 (because pH = -log [H+]). Therefore, using the power rule, we can re-express this as pH = log ([H+]^-1).

Using another log rule, we can express each side of this equation as an exponent of 10 and we get:

10^pH = 10^log ([H+]^-1).

Using a definition of logs, the right hand side of this equation now just becomes [H+]^-1. So we have:

10^pH = [H+]^-1

The right hand side can be expressed as 1/[H+] giving us:

10^pH = 1/[H+]

Multiplying each side by [H+] and dividing each side by 10^pH gives:

[H+] = 1/10^pH which is the same as saying [H+] = 10^-pH.

As an example, if the pH is 7, then [H+] = 10^-7.  