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 [H+][\text{H}^+] and pH\text{pH} using the following equations:
pH=log[H+][H+]=10pH\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ [\text H^+]&=10^{-\text{pH}}\end{aligned}
  • We can convert between [OH][\text{OH}^-] and pOH\text{pOH} using the following equations:
pOH=log[OH][OH]=10pOH\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ [\text {OH}^-]&=10^{-\text{pOH}}\end{aligned}
  • For any aqueous solution at 25C25\,^\circ\text{C}:
pH+pOH=14\text{pH}+\text{pOH}=14.
  • For every factor of 1010 increase in concentration of [H+][\text{H}^+], pH\text{pH} will decrease by 11 unit, and vice versa.
  • Both acid strength and concentration determine [H+][\text{H}^+] and pH\text{pH}.

Introduction

In aqueous solution, an acid is defined as any species that increases the concentration of H+(aq)\text{H}^+(aq), while a base increases the concentration of OH(aq)\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 25C25\,^\circ\text{C} contains 1.0×107 M1.0 \times 10^{-7}\text{ M} of H+\text{H}^+ and OH\text{OH}^-. In comparison, the concentration of H+{\text H^+} in stomach acid can be up to ~1.0×101M1.0 \times10^{-1}\,\text M. That means [H+][\text{H}^+] for stomach acid is approximately 66 orders of magnitude larger than in pure water!
To avoid dealing with such hairy numbers, scientists convert these concentrations to pH\text{pH} or pOH\text{pOH} values. Let's look at the definitions of pH\text{pH} and pOH\text{pOH}.

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

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

The pH\text{pH} for an aqueous solution is calculated from [H+][\text{H}^+] using the following equation:
pH=log[H+](Eq. 1a)\text{pH}=-\log[\text{H}^+]\quad\quad\quad\quad\;\;\;\text{(Eq. 1a)}
The lowercase p\text{p} indicates log10"``-\text{log}_{10}". You will often see people leave off the base 1010 part as an abbreviation.
For example, if we have a solution with [H+]=1×105 M[\text{H}^+]=1 \times 10^{-5}\text{ M}, then we can calculate the pH\text{pH} using Eq. 1a:
pH=log(1×105)=5.0\text{pH}=-\log(1 \times 10^{-5})=5.0
Given the pH\text{pH} of a solution, we can also find [H+][\text{H}^+]:
[H+]=10pH(Eq. 1b)[\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 pH\text {pH} and pOH\text {pOH}. For a review of log properties, see Sal's video on properties of logarithms.

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

The pOH\text{pOH} for an aqueous solution is defined in the same way for [OH][\text{OH}^-]:
pOH=log[OH] (Eq. 2a)\text{pOH}=-\log[\text{OH}^-]~\quad\quad\quad\;\;\text{(Eq. 2a)}
For example, if we have a solution with [OH]=1×1012 M[\text{OH}^-]=1 \times 10^{-12}\text{ M}, then we can calculate pOH\text{pOH} using Eq. 2a:
pOH=log(1×1012)=12.0\text{pOH}=-\log(1 \times 10^{-12})=12.0
Given the pOH\text{pOH} of a solution, we can also find [OH][\text{OH}^-]:
10pOH=[OH](Eq. 2b)10^{-\text{pOH}}=[\text {OH}^-]\quad\quad\quad\quad\quad\;\;\;\text{(Eq. 2b)}

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

Based on equilibrium concentrations of H+\text{H}^+ and OH\text{OH}^- in water, the following relationship is true for any aqueous solution at 25C25\,^\circ\text{C}:
pH+pOH=14  (Eq. 3)\text{pH}+\text{pOH}=14~~\quad\quad\quad\quad\quad\quad\quad\quad\text{(Eq. 3)}
This relationship can be used to convert between pH\text{pH} and pOH\text{pOH}. In combination with Eq. 1a/b and Eq. 2a/b, we can always relate pOH\text{pOH} and/or pH\text{pH} to [OH][\text{OH}^-] and [H+][\text{H}^+]. For a derivation of this equation, check out the article on the autoionization of water.

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

If we use 1.0 mmol1.0\text{ mmol} of NaOH\text{NaOH} to make 1.0 L1.0\text{ L} of an aqueous solution at 25C25\,^\circ\text{C}, what is the pH\text{pH} of this solution?
We can find the pH\text{pH} of our NaOH\text{NaOH} solution by using the relationship between [OH][\text{OH}^-], pH\text{pH}, and pOH\text{pOH}. Let's go through the calculation step-by-step.

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

Molar concentration is equal to moles of solute per liter of solution:
Molar concentration=mol soluteL solution\text{Molar concentration}=\dfrac{\text{mol solute}}{\text{L solution}}
To calculate the molar concentration of NaOH\text{NaOH}, we can use the known values for the moles of NaOH\text{NaOH} and the volume of solution:
[NaOH]=1.0 mmol NaOH1.0 L=1.0×103 mol NaOH1.0 L=1.0×103 M NaOH\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 NaOH\text{NaOH} in the solution is 1.0×103 M1.0\times10^{-3}\text{ M}.

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

Because NaOH\text{NaOH} is a strong base, it dissociates completely into its constituent ions in aqueous solution:
NaOH(aq)Na+(aq)+OH(aq)\text{NaOH}(aq)\rightarrow\text{Na}^+(aq)+\text{OH}^-(aq)
This balanced equation tells us that every mole of NaOH\text{NaOH} produces one mole of OH\text{OH}^- in aqueous solution. Therefore, we have the following relationship between [NaOH][\text{NaOH}] and [OH][\text{OH}^-]:
[NaOH]=[OH]=1.0×103 M[\text{NaOH}]=[\text{OH}^-]=1.0\times10^{-3}\text{ M}

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

Now that we know the concentration of OH\text{OH}^-, we can calculate pOH\text{pOH} using Eq. 2a:
pOH=log[OH]=log(1.0×103)=3.00\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ &=-\log(1.0\times10^{-3})\\ \\ &=3.00\end{aligned}
The pOH\text{pOH} of our solution is 3.003.00.

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

We can calculate pH\text{pH} from pOH\text{pOH} using Eq. 3. Rearranging to solve for our unknown, pH\text{pH}:
pH=14pOH\text{pH}=14-\text{pOH}
We can substitute the value of pOH\text{pOH} we found in Step 3 to find the pH\text{pH}:
pH=143.00=11.00\text{pH}=14-3.00=11.00
Therefore, the pH\text{pH} of our NaOH\text{NaOH} solution is 11.0011.00.

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

Converting [H+][\text{H}^+] to pH\text{pH} is a convenient way to gauge the relative acidity or basicity of a solution. The pH\text{pH} scale allows us to easily rank different substances by their pH\text{pH} value.
The pH\text{pH} scale is a negative logarithmic scale. The logarithmic part means that pH\text{pH} changes by 11 unit for every factor of 1010 change in concentration of H+\text{H}^+. The negative sign in front of the log tells us that there is an inverse relationship between pH\text{pH} and [H+][\text{H}^+]: when pH\text{pH} increases, [H+][\text{H}^+] decreases, and vice versa.
The following image shows a pH\text{pH} scale labeled with pH\text{pH} values for some common household substances. These pH\text{pH} values are for solutions at 25C25\,^\circ\text{C}. Note that it is possible to have a negative pH\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 25C25\,^\circ\text{C}:
  • For a neutral solution, pH=7\text{pH}=7.
  • Acidic solutions have pH<7\text{pH}<7.
  • Basic solutions have pH>7\text{pH}>7.
The lower the pH\text{pH} value, the more acidic the solution and the higher the concentration of H+\text H^+. The higher the pH\text{pH} value, the more basic the solution and the lower the concentration of H+\text H^+. While we could also describe the acidity or basicity of a solution in terms of pOH\text{pOH}, it is a little more common to use pH\text{pH}. Luckily, we can easily convert between pH\text{pH} and pOH\text{pOH} values.
Concept check: Based on the pH\text{pH} scale given above, which solution is more acidic-orange juice, or vinegar?
Vinegar has a pH\text{pH} of 3.0\approx3.0, and orange juice has a pH\text{pH} of 3.6\approx3.6. Because the pH\text{pH} of vinegar is less than that of orange juice (3.0<3.6)(3.0<3.6), vinegar has a higher concentration of H+\text{H}^+.
Therefore, vinegar is more acidic than orange juice.

Example 22: Determining the pH\text{pH} of a diluted strong acid solution

We have 100 mL100\text{ mL} of a nitric acid solution with a pH\text{pH} of 4.04.0. We dilute the solution by adding water to get a total volume of 1.0 L1.0\text{ L}.
What is the pH\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 pH\text{pH} scale is a negative logarithmic scale. Therefore, if the concentration of H+\text{H}^+ decreases by a single factor of 1010, then the pH\text{pH} will increase by 11 unit.
Since the original volume, 100 mL100\text{ mL}, is one tenth the total volume after dilution, the concentration of H+\text H^+ in solution has been reduced by a factor of 1010. Therefore, the pH\text{pH} of the solution will increase 11 unit:
pH=original pH+1.0=4.0+1.0=5.0\begin{aligned}\text{pH}&=\text{original pH}+1.0 \\ \\ &=4.0+1.0\\ \\ &=5.0\end{aligned}
Therefore, the pH\text{pH} of the diluted solution is 5.05.0.

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

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

We can use the pH\text{pH} and volume of the original solution to calculate the moles of H+\text{H}^+ in the solution.
moles H+=[H+]initial×volume=10pHM×volume=104.0M×0.100 L=1.0×105mol H+\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 H+\text{H}^+ after dilution

The molarity of the diluted solution can be calculated by using the moles of H+\text{H}^+ from the original solution and the total volume after dilution.
[H+]final=mol H+L solution=1.0×105mol H+1.0L=1.0×105M\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 pH\text{pH} from [H+][\text{H}^+]

Finally, we can use Eq. 1a to calculate pH\text{pH}:
pH=log[H+]=log(1.0×105)=5.0\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 pH\text{pH}. Method 1 is a handy shortcut when changes in concentration occur as multiples of 1010. Method 1 can also be used as a quick way to estimate pH\text{pH} changes.

Relationship between pH\text{pH} and acid strength

Based on the equation for pH\text{pH}, we know that pH\text{pH} is related to [H+][\text{H}^+]. However, it is important to remember that pH\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 [H+][\text{H}^+] at a given acid concentration. For example, a 1.0M1.0\,\text M solution of strong acid HCl\text{HCl} will have a higher concentration of H+\text{H}^+ than a 1.0M1.0\,\text M solution of weak acid HF\text{HF}. Thus, for two solutions of monoprotic acid at the same concentration, pH\text{pH} will be proportional to acid strength.
More generally though, both acid strength and concentration determine [H+][\text{H}^+]. Therefore, we can't always assume that the pH\text{pH} of a strong acid solution will be lower than the pH\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 [H+][\text{H}^+] and pH\text{pH} using the following equations:
pH=log[H+][H+]=10pH\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ [\text H^+]&=10^{-\text{pH}}\end{aligned}
  • We can convert between [OH][\text{OH}^-] and pOH\text{pOH} using the following equations:
pOH=log[OH][OH]=10pOH\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ [\text {OH}^-]&=10^{-\text{pOH}}\end{aligned}
  • For every factor of 1010 increase in concentration of [H+][\text{H}^+], pH\text{pH} will decrease by 11 unit, and vice versa.
  • For any aqueous solution at 25C25\,^\circ\text{C}:
pH+pOH=14\text{pH}+\text{pOH}=14.
  • Both acid strength and concentration determine [H+][\text{H}^+] and pH\text{pH}.

Attributions

This article was adapted from the following articles:
  1. The pH Scale” from UC Davis ChemWiki, CC BY-NC-SA 3.0
The modified article is licensed under a CC-BY-NC-SA 4.0 license.

Additional References

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 11: Calculating the pH\text{pH} of a strong base solution at 25C25\,^\circ\text{C}

We make 200 mL200\text{ mL} of a solution with a 0.025 M0.025\text{ M} concentration of Ca(OH)2\text{Ca(OH)}_2. The solution is then diluted to 1.00 L1.00\text{ L} by adding additional water.
What is the pH\text{pH} of the solution after dilution?
Choose 1 answer:
Choose 1 answer:
Calcium hydroxide dissociates in aqueous solution as follows:
Ca(OH)2(aq)Ca2+(aq)+2OH(aq)\text{Ca(OH)}_2(aq)\rightarrow\text{Ca}^{2+}(aq)+2\text{OH}^-(aq)
We see that 2 mol OH2\text{ mol OH}^- are produced for every 1 mol Ca(OH)21\text{ mol Ca(OH)}_2 that dissociates. Therefore, the concentration of hydroxide ion, [OH][\text{OH}^-], will be twice the molarity of Ca(OH)2\text{Ca(OH)}_2:
[OH]initial=0.025 M×2mol OH1mol Ca(OH)2=0.050 M[\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 55, since 200 mL×5=1.0 L200\text{ mL}\times5=1.0\text{ L}. Therefore, the final molar concentration of OH\text{OH}^- will be 15\dfrac{1}{5} its initial value:
[OH]final=0.050 M×15=0.010 M[\text{OH}^-]_\text{final}=0.050\text{ M}\times\dfrac{1}{5}=0.010\text{ M}
We can calculate the pOH\text{pOH} of the diluted solution from [OH][\text{OH}^-]:
pOH=log[OH]=log(0.010)=2.00\text{pOH}=-\log[\text{OH}^-]=-\log(0.010)=2.00
Since pH+pOH=14.00\text{pH}+\text{pOH}=14.00 at 25C25\,^\circ\text{C},
pH=14.00pOH=14.002.00=12.00\begin{aligned}\text{pH}&=14.00-\text{pOH}\\ \\ &=14.00-2.00\\ \\ &=12.00\end{aligned}
Therefore, the pH\text{pH} of the diluted solution is 12.0012.00.
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