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E3.5 · Use water ionization to calculate pH, pOH, and ion concentrations

Learn to use water ionization to calculate ph, poh, and ion concentrations through clear examples and targeted practice.

Ontario Grade 12 Chemistry

Chemical Systems and Equilibrium

SCH4U study topic E3.5

A solution can be acidic, neutral, or basic, but its appearance alone does not tell us the concentrations of the ions responsible. A quantitative scale called pH helps describe acidity. To use it, connect the particle model of water to the concentrations of hydronium and hydroxide ions. You will then use a small set of relationships to calculate pH, pOH, or either ion concentration. This lesson uses concentrations in mol/L and the water ionization constant at 25 °C.

What you will learn

  • Describe how water forms hydronium and hydroxide ions.
  • Use the water ionization constant to relate hydronium and hydroxide concentrations at 25 °C.
  • Calculate pH, pOH, and ion concentrations, and report results with appropriate significant digits.

1. From water particles to ions

You already know that a solution contains particles, and that concentration describes the amount of a dissolved substance in a given volume. For these calculations, ion concentration means the amount of a particular ion per litre of solution. It is written in mol/L.
Even pure water contains tiny amounts of ions. In a particle-level model, one water molecule transfers a hydrogen ion to another water molecule. The products are a hydronium ion and a hydroxide ion. Hydronium is written as H₃O⁺, and hydroxide is written as OH⁻. The positive and negative charges are equal in size, so the ions formed by water ionization balance each other in pure water.
This process is called water ionization. It does not mean that all water molecules become ions. Most remain as H₂O molecules. The reaction shows the small amount of ion formation that matters for pH calculations.
2H2O(l)⇌H3O+(aq)+OH−(aq)2\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)}
  • Hydronium ions have the formula H₃O⁺.
  • Hydroxide ions have the formula OH⁻.
  • Pure water produces equal concentrations of these two ions.

2. The concentration relationship and pH scale

At 25 °C, the product of the hydronium and hydroxide concentrations in an aqueous solution is the water ionization constant. In this course-level calculation, use its value as 1.0 × 10⁻¹⁴. The square brackets mean “concentration of.” Thus, [H₃O⁺] means the concentration of hydronium ions, in mol/L.
This relationship means the two ion concentrations are linked. If the hydronium concentration is known, the hydroxide concentration can be found by dividing the constant by the hydronium concentration. The same rearrangement works in reverse. In pure water, the two concentrations are equal, so each is 1.0 × 10⁻⁷ mol/L at 25 °C.
pH is a number that describes hydronium concentration. pOH describes hydroxide concentration. Both are calculated using the negative base-10 logarithm of the relevant concentration. A logarithm is the operation that answers the question, “To what power must 10 be raised to give this number?” For example, 10⁻³ has a base-10 logarithm of −3.
At 25 °C, a pH below 7 indicates an acidic solution, pH 7 indicates a neutral solution, and pH above 7 indicates a basic solution. These comparisons are useful, but calculations should begin with the stated concentration or pH rather than assuming a solution’s type.
Kw=[H3O+][OH−]=1.0×10−14K_w=[\mathrm{H_3O^+}][\mathrm{OH^-}]=1.0\times10^{-14}
  • Use the water ionization constant at 25 °C.
  • The pH and pOH equations use concentrations in mol/L.
  • At 25 °C, pH and pOH add to 14.00.

3. Calculating pH, pOH, and concentrations

Choose the relationship that matches the quantity provided. If an ion concentration is given, use the corresponding logarithm equation to find pH or pOH. If a pH or pOH is given, use a power of 10 to recover the concentration. If one ion concentration is given and the other is needed, use the water ionization constant.
The concentration form of the pH equation is [H₃O⁺] = 10⁻ᵖᴴ mol/L. The matching pOH relationship is [OH⁻] = 10⁻ᵖᴼᴴ mol/L. These are rearrangements of the logarithm definitions, not new chemical reactions.
At 25 °C, pH + pOH = 14.00. This gives a quick route from one value to the other. For example, once pH is known, subtract it from 14.00 to find pOH. Then use the pOH relationship to find hydroxide concentration if needed.
Keep concentration units visible during the calculation. Logarithm results such as pH and pOH are reported without units. For logarithms, the digits after the decimal point in the pH or pOH correspond to the significant figures in the concentration. For instance, a concentration recorded with two significant figures gives a pH reported to two decimal places. Avoid rounding intermediate values; round the final result to suit the given data.
pH=−log⁡[H3O+],pOH=−log⁡[OH−],pH+pOH=14.00\mathrm{pH}=-\log[\mathrm{H_3O^+}],\quad \mathrm{pOH}=-\log[\mathrm{OH^-}],\quad \mathrm{pH}+\mathrm{pOH}=14.00
  • Use the logarithm equation for concentration-to-pH calculations.
  • Use powers of 10 for pH-to-concentration calculations.
  • For this course-level treatment at 25 °C, pH + pOH = 14.00.

4. A reliable calculation routine

Start by listing what is given and what is requested. Identify whether the given quantity is a concentration, pH, or pOH. Then select one equation that directly connects the known and unknown quantities.
Check that concentrations are expressed in mol/L before substituting. If both ion concentrations are involved, check that their product matches the water ionization constant. Finally, check whether the answer makes sense: a larger hydronium concentration corresponds to a lower pH, while a larger hydroxide concentration corresponds to a lower pOH.
Keep the temperature condition in mind. The value used here for the water ionization constant, and the pH-plus-pOH relationship, are for 25 °C. Do not treat these as unrelated memorized rules: both follow from the water ionization relationship used at that temperature.
  • Identify the known quantity before choosing an equation.
  • Check units and significant figures.
  • Use 25 °C conditions for the stated constant and pH-plus-pOH relationship.

Worked example

Find pH and both ion concentrations

A solution at 25 °C has a hydronium concentration of 3.2 × 10⁻⁵ mol/L. Calculate its pH, pOH, and hydroxide concentration. Report the pH and pOH to two decimal places, and the concentrations to two significant figures.
  1. Calculate pH
    Use the hydronium concentration in the pH definition. The given concentration has two significant figures, so report pH to two decimal places.
    pH=−log⁡(3.2×10−5)=4.49\mathrm{pH}=-\log(3.2\times10^{-5})=4.49
  2. Calculate pOH
    The solution is at 25 °C, so use the relationship between pH and pOH. Subtract the calculated pH from 14.00.
    pOH=14.00−4.49=9.51\mathrm{pOH}=14.00-4.49=9.51
  3. Calculate hydroxide concentration
    Rearrange the water ionization relationship to find hydroxide concentration. Divide the constant by the given hydronium concentration. The concentration is reported to two significant figures.
    [OH−]=1.0×10−143.2×10−5 mol/L=3.1×10−10 mol/L[\mathrm{OH^-}]=\frac{1.0\times10^{-14}}{3.2\times10^{-5}\ \mathrm{mol/L}}=3.1\times10^{-10}\ \mathrm{mol/L}
Answer: pH = 4.49; pOH = 9.51; [OH⁻] = 3.1 × 10⁻¹⁰ mol/L.
Check: The ion concentrations multiply to approximately 1.0 × 10⁻¹⁴. The pH is below 7, and the hydroxide concentration is less than the hydronium concentration, which is consistent with the calculated values.

Common mistakes and how to avoid them

Using the hydroxide concentration in the pH equation.
Correction: Use hydronium concentration for pH and hydroxide concentration for pOH.
Multiplying by the water ionization constant to find the second ion concentration.
Correction: Rearrange the product relationship by dividing the constant by the known ion concentration.
Reporting pH with units such as mol/L.
Correction: Concentrations have units of mol/L. pH and pOH are reported as numbers without units.
Treating pH 7 as neutral at any temperature.
Correction: This lesson’s neutral-water comparison and pH-plus-pOH relationship use the stated 25 °C conditions.

Lesson summary

  • Water ionization forms hydronium and hydroxide ions.
  • At 25 °C, the product of their concentrations is 1.0 × 10⁻¹⁴.
  • pH describes hydronium concentration; pOH describes hydroxide concentration.
  • Use pH + pOH = 14.00 at 25 °C and report concentrations in mol/L.

Check your understanding

Question 1

At 25 °C, a solution has pH 3.62. What is its pOH?
  1. 10.38
  2. 3.62
  3. 14.00
  4. 7.00
Show answer and explanation
10.38
Subtract pH from 14.00: pOH = 14.00 − 3.62 = 10.38.

Question 2

At 25 °C, [H₃O⁺] = 2.5 × 10⁻⁴ mol/L. What is [OH⁻]?
  1. 4.0 × 10⁻¹¹ mol/L
  2. 2.5 × 10⁻¹⁸ mol/L
  3. 4.0 × 10⁻⁴ mol/L
  4. 2.5 × 10⁻⁴ mol/L
Show answer and explanation
4.0 × 10⁻¹¹ mol/L
Divide 1.0 × 10⁻¹⁴ by 2.5 × 10⁻⁴ mol/L. The result is 4.0 × 10⁻¹¹ mol/L to two significant figures.

Question 3

A solution at 25 °C has pH 9.30. What is its hydronium concentration?
  1. 5.0 × 10⁻¹⁰ mol/L
  2. 9.30 mol/L
  3. 2.0 × 10⁻⁵ mol/L
  4. 1.0 × 10⁻⁹ mol/L
Show answer and explanation
5.0 × 10⁻¹⁰ mol/L
Use [H₃O⁺] = 10⁻ᵖᴴ mol/L. The concentration is 10⁻⁹·³⁰ mol/L, or 5.0 × 10⁻¹⁰ mol/L to two significant figures.

Key terms

Concentration
The amount of a substance in a stated volume; here, ion concentration is expressed in mol/L.
Hydronium ion
The positively charged ion H₃O⁺ formed when a water molecule receives a hydrogen ion.
Hydroxide ion
The negatively charged ion OH⁻ formed during water ionization.
Water ionization constant
The product of hydronium and hydroxide concentrations in water at a stated temperature.
pH
A number calculated from hydronium concentration that describes how acidic a solution is.
pOH
A number calculated from hydroxide concentration.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Chemistry (SCH4U), expectation E3.5. It is a study resource, not an official curriculum publication.

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