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Acids and Bases - The pH scale

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The pHpH scale is a logarithmic scale used to represent the concentration of hydrogen ions [H+][H^+] in an aqueous solution. It typically ranges from 00 to 1414 at 298 K298\text{ K}.

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pHpH is defined as the negative logarithm to the base 1010 of the hydrogen ion concentration: pH=−log⁡10[H+]pH = -\log_{10}[H^+].

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A change of one pHpH unit represents a ten-fold change in the concentration of hydrogen ions [H+][H^+]. For example, a solution with pH=2pH = 2 has 1010 times the [H+][H^+] of a solution with pH=3pH = 3.

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The ionic product constant for water, KwK_w, is defined as Kw=[H+][OH−]K_w = [H^+][OH^-]. At 298 K298\text{ K}, Kw=1.0×10−14K_w = 1.0 \times 10^{-14}.

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In any aqueous solution at 298 K298\text{ K}, the relationship between pHpH and pOHpOH is pH+pOH=pKw=14.00pH + pOH = pK_w = 14.00.

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In a neutral solution, [H+]=[OH−][H^+] = [OH^-]. At 298 K298\text{ K}, this corresponds to pH=7.0pH = 7.0. Acidic solutions have pH<7pH < 7 and basic (alkaline) solutions have pH>7pH > 7.

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The value of KwK_w is temperature-dependent. Since the auto-ionization of water (H2O(l)⇌H+(aq)+OH−(aq)H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)) is endothermic, KwK_w increases as temperature increases, which means the pHpH of pure water decreases as temperature rises, even though it remains neutral ([H+]=[OH−][H^+] = [OH^-]).

📐Formulae

pH=−log⁡10[H+(aq)]pH = -\log_{10}[H^+(aq)]

[H+]=10−pH[H^+] = 10^{-pH}

pOH=−log⁡10[OH−(aq)]pOH = -\log_{10}[OH^-(aq)]

[OH−]=10−pOH[OH^-] = 10^{-pOH}

Kw=[H+][OH−]K_w = [H^+][OH^-]

pKw=−log⁡10KwpK_w = -\log_{10}K_w

pH+pOH=pKwpH + pOH = pK_w

pH+pOH=14.00 (at 298 K)pH + pOH = 14.00 \text{ (at } 298\text{ K)}

💡Examples

Problem 1:

Calculate the pHpH of a 0.050 mol dm−30.050 \text{ mol dm}^{-3} solution of barium hydroxide, Ba(OH)2Ba(OH)_2, at 298 K298\text{ K}.

Solution:

[OH−]=2×0.050=0.10 mol dm−3[OH^-] = 2 \times 0.050 = 0.10 \text{ mol dm}^{-3} pOH=−log⁡10(0.10)=1.00pOH = -\log_{10}(0.10) = 1.00 pH=14.00−1.00=13.00pH = 14.00 - 1.00 = 13.00

Explanation:

Ba(OH)2Ba(OH)_2 is a strong base that dissociates completely to release two OH−OH^- ions per formula unit. Therefore, the concentration of hydroxide ions is twice the concentration of the base. We find pOHpOH first and then subtract from 1414 to find pHpH.

Problem 2:

If the pHpH of a solution is 4.704.70, calculate the concentration of hydrogen ions [H+][H^+] in mol dm−3\text{mol dm}^{-3}.

Solution:

[H+]=10−pH[H^+] = 10^{-pH} [H+]=10−4.70[H^+] = 10^{-4.70} [H+]=2.00×10−5 mol dm−3[H^+] = 2.00 \times 10^{-5} \text{ mol dm}^{-3}

Explanation:

To convert pHpH back to concentration, use the antilog function (base 1010). In IB Chemistry, ensure the final answer is given to the appropriate number of significant figures.

Problem 3:

At a higher temperature, the KwK_w of water is 5.48×10−145.48 \times 10^{-14}. Calculate the pHpH of pure water at this temperature and state whether the water is acidic, basic, or neutral.

Solution:

[H+]=Kw=5.48×10−14=2.34×10−7 mol dm−3[H^+] = \sqrt{K_w} = \sqrt{5.48 \times 10^{-14}} = 2.34 \times 10^{-7} \text{ mol dm}^{-3} pH=−log⁡10(2.34×10−7)=6.63pH = -\log_{10}(2.34 \times 10^{-7}) = 6.63 The water is neutral.

Explanation:

In pure water, [H+]=[OH−][H^+] = [OH^-], so Kw=[H+]2K_w = [H^+]^2. Although the pHpH is less than 77, the water is still neutral because the concentration of H+H^+ ions still equals the concentration of OH−OH^- ions.

The pH scale - Revision Notes & Key Formulas | IB Grade 12 Chemistry