Lesson Explainer: Calculating pH Chemistry

In this explainer, we will learn how to calculate the pH of a solution and describe the ion-product constant of water.

Water is an amphoteric substance. This means that a molecule of water can behave as an acid or as a base. If water can behave as an acid or as a base, this means that one water molecule can undergo an acid–base reaction with another water molecule. In this acid–base reaction, the acidic water molecule will lose a proton and the basic water molecule will gain a proton.

The reaction shown above is known as the autoionization of water.

Reaction: Autoionization of Water

The chemical equation for the acid–base equilibrium reaction of two water molecules is HO()+HO()HO()+OH()223+–llaqaq

This reaction occurs in every water sample, but as water is a weak acid and a weak base, only a fraction of the water molecules participate in the reaction.

The autoionization of water is an equilibrium reaction, where the forward and reverse reactions occur at the same rate. This means that the concentration of the reactants and products will remain constant even though both reactions are still occurring. For any reaction at equilibrium, we can determine an equilibrium constant (𝐾) from the equilibrium expression.

Definition: Equilibrium Constant (𝐾𝑐)

Equilibrium Constant (𝐾) is a value that expresses the relationship between products and reactants at a given temperature.

For the reaction 𝑎+𝑏𝑐+𝑑ABCD where lowercase letters represent molar coefficients and capital letters represent chemical symbols, the equilibrium expression would be 𝐾=.[C][D][A][B]

The brackets in the equation indicate that we must use the concentration of the species when solving the expression. The concentration should have units of molarity (M).

We can apply this idea to the autoionization of water: HO+HOHO+OH223+–

The equilibrium expression for the autoionization of water would be 𝐾=.[HO][OH][HO][HO]3+–22

As only a very small amount of water reacts to produce hydroxide and hydronium, the concentration of water remains virtually unchanged and can be omitted from the equation. This leaves us with the following equation: 𝐾=.w3+–[HO][OH]

The 𝐾 is replaced by 𝐾w to indicate that this is the equilibrium constant for the autoionization of water. This is also known as the ion-product constant.

Definition: Ion-Product Constant of Water (𝐾w)

The ion-product constant of water (𝐾)w is the equilibrium constant for the autoionization of water. It is expressed by the equation 𝐾=.w3+–[HO][OH]

It is important to recognize that hydronium (HO3+) and hydrogen ions (H+) are often used interchangeably in acid–base equations. Thus, the ion-product constant expression may also be written as 𝐾=.w+–[H][OH]

In a sample of pure water at 25∘C, the concentrations of hydronium and hydroxide are both 1.0×10 M. Substituting these concentrations into the ion-product constant expression gives a value of 1.0×10 for 𝐾w.

Constant: Ion-Product Constant (𝐾w) at 25°C

𝐾==1.0×10w3+–[HO][OH]

As this is the equilibrium constant for water at 25∘C, this expression holds true for any aqueous solution at 25∘C. Notice that the values of 𝐾w, [HO3+], and [OH–] are dependent on the temperature of the solution. Changing the temperature will change the value of 𝐾w and the concentration of both ions.

The concentration of hydronium and hydroxide determines whether the solution is acidic, basic, or neutral. The relationships between these ions are shown in the table below.

[HO][OH]3+–Neutral
[HO]>[OH]3+–Acidic
[HO]<[OH]3+–Basic

Example 1: Calculating and Comparing the Concentration of Hydroxide from the Proton Concentration of an Aqueous Solution

A solution at 25∘C has an [OH]– of 1.26×10 mol/L. Is the solution acid, basic, or neutral?

Answer

In order to determine if a solution is acidic, basic, or neutral, we will need to compare the concentration of hydroxide ions, [OH]–, to the concentration of hydronium ions, [HO]3+. At 25∘C, the ion-product constant (𝐾)w of a solution will be 1.0×10; thus, the following equation can be used to relate [OH]– and [HO]3+: 𝐾=1×10=.w3+–[HO][OH]

We can substitute the hydroxide ion concentration into the equation: 1×10=×1.26×10.[HO]3+

We divide both sides of the equation by the hydroxide ion concentration to determine the hydronium ion concentration: 1×10(1.26×10)=×1.26×10(1.26×10)7.94×10=.[HO]molL[HO]3+3+

We should recognize that although the ion-product constant is unitless, the concentration of hydronium ions should be given the unit of molarity (M or mol/L).

Now we can compare [OH]– and [HO]3+. The hydroxide ion concentration is less than the hydronium ion concentration: =1.26×10/<7.94×10/=.[OH]molLmolL[HO]–3+

At 25∘C, the following relationships are observed:

[HO][OH]3+–Neutral
[HO]>[OH]3+–Acidic
[HO]<[OH]3+–Basic

As the concentration of hydroxide ions in the solution is less than the concentration of hydronium ions, the solution is acidic.

Example 2: Calculating the Hydroxide Concentration of an Aqueous Solution given the Proton Concentration

An aqueous solution at 25∘C has a [H]+ of 2×10 M. What is the value of [OH]– to the nearest whole number?

Answer

At 25∘C, the ion-product constant (𝐾)w can be expressed by the following equation: 𝐾==1.0×10.w+–[H][OH]

We can substitute the proton concentration, [H]+, into the equation and rearrange to solve for the hydroxide concentration, [OH]–: 2×10×=1.0×102×10×(2×10)=1.0×10(2×10)=5×10.[OH][OH][OH]–––

We must recognize that the brackets indicate that we have solved for a concentration. This means that our answer should be given the unit mol/L or M. The value of [OH]– is then 5×10 M.

The concentrations of hydroxide and hydronium in a solution are very small and frequently written in scientific notation. In 1909, the Danish chemist Soren Sorensen proposed a more practical method for representing the concentration of hydronium, pH. pH is a unitless value determined by taking the negative logarithm of the hydrogen ion concentration.

Equation: pH of a Solution

The pH of a solution can be calculated using the following equation: pHlog[HO]orpHlog[H]=−=−.3++

The purpose of the logarithm in the equation is to produce a value that is more convenient to work with. The result of a base 10 logarithm, as used in the pH equation, indicates how many times the number one must be multiplied or divided by 10 to give the desired value.

For example, one must be multiplied by 10 two times to give the value 100. Thus, the logarithm of 100 is two: 1×10×10=100;,(100)=2.thereforelog

One must be divided by 10 three times to give the value 0.001. Thus, the logarithm of 0.001 is negative three. The negative indicates that we are dividing by 10 rather than multiplying: 110×10×10=0.001;,(0.001)=−3.thereforelog

Values that are not exact multiples of 10 will not have whole number logarithms. Examples of several base 10 logarithms are shown in the table below.

log1.0×10=−8
log0.001=−3
log0.05=−1.3
log0.1=−1
log1=0
log100=2
log500=2.7
log1000=3
log1.0×10=5

For pure water at 25∘C, the concentration of hydronium is 1.0×10 M. This means that the pH of pure water at 25∘C is 7: pHlog[HO]log=−=−1.0×10=7.3+

In the following table, the calculated pHs of several aqueous solutions at 25∘C are shown.

[HO]3+[OH]–Acidic, Basic, or NeutralpHlog[HO]=−3+
1.0×10 M1.0×10 MBasic12.00
2.7×10 M3.7×10 MBasic8.57
1.0×10 M1.0×10 MNeutral7.00
8.4×10 M1.2×10 MAcidic4.08
1.0×10 M1.0×10 MAcidic3.00

We stated earlier that a solution is neutral when [HO][OH]3+–, acidic when [HO]>[OH]3+–, and basic when [HO]<[OH]3+–. Notice that solutions that are acidic have pH values less than 7, solutions that are basic have pH values greater than 7, and neutral solutions have a pH value equal to 7. The pH scale with common acids and bases is shown below.

Example 3: Calculating the pH of a Solution with a Given Hydronium Ion Concentration

What is the pH of a solution with a [HO]3+ of 3.98×10 mol⋅dm−3? Give your answer to 1 decimal place.

Answer

Remember that the notation [HO]3+ represents the concentration of hydronium ions in the solution. We can use the concentration to calculate the pH via the following equation: pHlog[HO]=−,3+ where the concentration of hydronium ions should have a unit of molarity (M). The concentration was given in mol⋅dm−3. Notice that the following units are equal: moldmmoldmmolLM⋅===.

Therefore, 3.98×10 mol⋅dm−3 can be written with the unit of molarity, 3.98×10 M, and can be substituted into the pH equation: pHlogpH=−3.98×10=1.40012.

Giving our answer to one decimal place, the pH of the solution is 1.4.

Often, we can measure the pH of a solution directly with a pH meter, but we can calculate the pH from the hydronium concentration of the solution.

Equation: Hydronium Ion Concentration of a Solution

The hydronium ion concentration of a solution can be calculated using the following equation: [HO]or[H]3+pH+pH=10=10.

Example 4: Calculating the Hydronium Ion Concentration of a Solution from a Given pH

The pH of a solution is measured to be 4.15. What is the concentration of hydronium ions in this solution? Give your answer to 2 decimal places and in units of mol⋅dm−3.

Answer

The concentration of hydronium ions can be represented with the symbol [HO]3+. We can use the pH to calculate the hydronium ion concentration via the following equation: [HO]3+pH=10.

Substituting the pH into the equation gives [HO][HO]3+3+=10=7.07946×10.

The numerical value of the hydronium ion concentration is 7.07946×10.

When using this equation, recognize that the unit for the concentration of hydronium ions will be molarity (M). However, the question asks for the answer to have the unit mol⋅dm−3. We should recognize that the following units are equal: MmolLmoldmmoldm===⋅.

Rounding our answer to 2 decimal places, we have determined that the concentration of hydronium ions in the solution is 7.08×10 mol⋅dm−3.

Just as the concentration of hydronium ions can be used to calculate the pH, the concentration of hydroxide ions can be used to calculate the pOH.

Equation: pOH of a Solution

The pOH of a solution can be calculated using the following equation: pOHlog[OH]=−.–

The pOH values for each aqueous solution at 25∘C have been added to the table below.

[HO]3+[OH]–Acidic, Basic, or NeutralpHlog[HO]=−3+pOHlog[OH]=−–
1.0×10 M1.0×10 MBasic122
2.7×10 M3.7×10 MBasic8.575.43
1.0×10 M1.0×10 MNeutral77
8.4×10 M1.2×10 MAcidic4.089.92
1.0×10 M1.0×10 MAcidic311

Notice that the pOH of a basic solution is less than 7. For neutral solutions, the pOH is equal to 7. For acidic solutions, the pOH is greater than 7. This pOH trend is the exact opposite of the pH trend. We should also be able to notice that the sum of the pH and pOH is 14, which allows us to easily relate them.

Equation: Relationship between pH and pOH at 25°C

The pH and pOH of a solution at 25∘C can be related using the following equation: pHpOH+=14.

The pH can be equated to the hydronium concentration. Likewise, the pOH can be used to determine the hydroxide concentration.

Equation: Hydroxide Ion Concentration of a Solution

The hydroxide ion concentration of a solution can be calculated using the following equation: [OH]–pOH=10.

Altogether, pH, pOH, [HO]3+, and [OH]– can be related to one another via six different equations. If just one of the four quantities is known, the remaining three can be determined. The relationships between these quantities is represented by the diagram below.

Example 5: Calculating the pH of a Solution from the Hydroxide Ion Concentration

A solution at 25∘C has a hydroxide ion concentration of 1.26×10 mol⋅dm−3. What is the pH of this solution to one decimal place?

Answer

First, we must recognize that hydroxide ion concentration can be represented by the symbol [OH]–. We should also recognize that the following units can be used interchangeably to represent the concentration: moldmmoldmmolLM⋅===.

Thus, we know that [OH]M–=1.26×10. In order to calculate the pH, we must either determine the hydronium concentration or the pOH. Both possible methods for solving will be shown below.

Method One

In this method, we first determine the hydronium concentration. Hydroxide and hydronium are related by the following equation when the solution is at 25∘C: [HO][OH]3+–=1.0×10.

As the solution in the question is at 25∘C, we can substitute the hydroxide concentration into the equation and solve for the hydronium concentration: [HO][HO][HO]M3+3+3+×1.26×10=1.0×10×1.26×10(1.26×10)=1.0×10(1.26×10)=7.94×10.

The hydronium ion concentration is 7.94×10 M. We can substitute this value into the pH equation to solve for the pH of the solution: pHlog[HO]pHlogpH=−=−7.94×10=5.1.3+

Method Two

In this method, we first determine the pOH of the solution by substituting the hydroxide concentration into the pOH equation: pOHlog[OH]pOHlogpOH=−=−1.26×10=8.9.–

pOH is related to pH at 25∘C by the following equation: pHpOH+=14.

As the solution is at 25∘C, we can substitute the pOH into the equation and solve for the pH: pHpH+8.9=14=5.1.

Both methods have given us a pH to one decimal place of 5.1.

Key Points

  • The autoionization of water is an acid–base equilibrium reaction between two water molecules.
  • The autoionization of water can be expressed by the ion–product constant.
  • pH is a representation of the concentration of hydronium ions in solution while pOH is a representation of the concentration of hydroxide ions in solution.
  • Solutions with a pH=7 are neutral, those with a pH<7 are acidic, and those with a pH>7 are basic.
  • Solutions with a pOH=7 are neutral, those with a pOH<7 are basic, and those with a pOH>7 are acidic.
  • pH, pOH, [HO]3+, and [OH]– can be related to one another using six equations: pHlog[HO][HO]pOHlog[OH][OH]pHpOH[HO][OH]=−,=10,=−,=10,+=14,𝐾=×=1.0×10.3+3+pH––pOHw3+–

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