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Difficulty: Very hardpH and pOH Scale and Calculations

A solution is formed by mixing 300 cm3300\text{ cm}^3 of 0.10 mol dm−30.10\text{ mol dm}^{-3} tetraoxosulfate(VI) acid (H2SO4\text{H}_2\text{SO}_4) solution with 700 cm3700\text{ cm}^3 of 0.10 mol dm−30.10\text{ mol dm}^{-3} sodium hydroxide (NaOH\text{NaOH}) solution at 25∘C25^\circ\text{C}. Assuming complete dissociation of both electrolytes, what is the pH of the resulting mixture?

Answer: 12

Answer

The pH of the resulting mixture is 12.0.
The mixture contains excess hydroxide ions (0.010 mol in 1.0 dm³ solution), resulting in a pOH of 2.0. Subtracting this from 14.0 gives a pH of 12.0.

Step-by-Step Solution

1
Calculate the moles of hydrogen ions (H⁺) contributed by the acid solution.
n(H+)=0.060 moln(\text{H}^+) = 0.060\text{ mol}
Tetraoxosulfate(VI) acid is diprotic (dibasic), releasing 2 moles of H+\text{H}^+ per mole of acid: 0.300 dm3×0.10 mol dm−3×2=0.060 mol0.300\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} \times 2 = 0.060\text{ mol}.
2
Calculate the moles of hydroxide ions (OH⁻) contributed by the base solution.
n(OH−)=0.070 moln(\text{OH}^-) = 0.070\text{ mol}
Sodium hydroxide is a monobasic base: 0.700 dm3×0.10 mol dm−3=0.070 mol0.700\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} = 0.070\text{ mol}.
3
Determine the unneutralized excess ions and calculate their molar concentration in the total volume.
[OH−]=0.010 mol dm−3[\text{OH}^-] = 0.010\text{ mol dm}^{-3}
The neutralization reaction is H++OH−→H2O\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}. Excess OH−=0.070−0.060=0.010 mol\text{OH}^- = 0.070 - 0.060 = 0.010\text{ mol}. Divided by the total mixture volume of 1.0 dm31.0\text{ dm}^3, [OH−]=0.010 mol dm−3[\text{OH}^-] = 0.010\text{ mol dm}^{-3}.
4
Calculate pOH and convert it to pH using the water autoionization relation.
pH=12.0\text{pH} = 12.0
pOH=−log⁡10(0.010)=2.0\text{pOH} = -\log_{10}(0.010) = 2.0. Since pH+pOH=14.0\text{pH} + \text{pOH} = 14.0 at 25∘C25^\circ\text{C}, pH=14.0−2.0=12.0\text{pH} = 14.0 - 2.0 = 12.0.

Key Concept

Neutralization stoichiometry of diprotic acids and strong bases followed by pH determination from excess hydroxide concentration.
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Topics

pH and pOH Scale and Calculations

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