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Acids and Bases - The pH scale

Grade 11IBChemistry

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

🔑Concepts

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The pHpH scale is a logarithmic representation of the concentration of hydrogen ions in an aqueous solution: pH=−log⁡10[H+(aq)]pH = -\log_{10}[H^+(aq)].

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Because the scale is logarithmic, a change of one pHpH unit corresponds to a tenfold (10×10\times) change in the hydrogen ion concentration [H+][H^+].

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The pOHpOH scale measures hydroxide ion concentration: pOH=−log⁡10[OH−]pOH = -\log_{10}[OH^-].

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

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The relationship between pHpH and pOHpOH at 298 K298\text{ K} is pH+pOH=14.00pH + pOH = 14.00.

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A neutral solution at 298 K298\text{ K} has [H+]=[OH−]=1.0×10−7 mol dm−3[H^+] = [OH^-] = 1.0 \times 10^{-7}\text{ mol dm}^{-3}, resulting in a pHpH of 7.07.0.

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Strong acids and strong bases are assumed to dissociate completely in aqueous solution to determine the initial [H+][H^+] or [OH−][OH^-] concentrations.

📐Formulae

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

[H+]=10−pH[H^+] = 10^{-pH}

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

Kw=[H+][OH−]=1.0×10−14 (at 298 K)K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ (at } 298\text{ K)}

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 nitric acid, HNO3(aq)HNO_3(aq).

Solution:

[H+]=0.050 mol dm−3[H^+] = 0.050\text{ mol dm}^{-3}. pH=−log⁡10(0.050)=1.30pH = -\log_{10}(0.050) = 1.30.

Explanation:

Nitric acid is a strong monoprotic acid, so it dissociates completely: HNO3→H++NO3−HNO_3 \rightarrow H^+ + NO_3^-. Therefore, the concentration of H+H^+ is equal to the concentration of the acid.

Problem 2:

Calculate the hydrogen ion concentration, [H+][H^+], for a solution with a pHpH of 4.754.75.

Solution:

[H+]=10−4.75=1.78×10−5 mol dm−3[H^+] = 10^{-4.75} = 1.78 \times 10^{-5}\text{ mol dm}^{-3}.

Explanation:

To find the concentration from pHpH, we use the inverse log formula [H+]=10−pH[H^+] = 10^{-pH}.

Problem 3:

Calculate the pHpH of a 0.010 mol dm−30.010\text{ mol dm}^{-3} solution of Ba(OH)2Ba(OH)_2 at 298 K298\text{ K}.

Solution:

[OH−]=2×0.010=0.020 mol dm−3[OH^-] = 2 \times 0.010 = 0.020\text{ mol dm}^{-3}. pOH=−log⁡10(0.020)=1.70pOH = -\log_{10}(0.020) = 1.70. pH=14.00−1.70=12.30pH = 14.00 - 1.70 = 12.30.

Explanation:

Ba(OH)2Ba(OH)_2 is a strong base that dissociates to give two OH−OH^- ions. We calculate pOHpOH first, then use the relationship pH+pOH=14pH + pOH = 14 to find the pHpH.