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Difficulty: HardpH and pOH Scale and Calculations

A sample of 0.62 g0.62\text{ g} of sodium oxide (Na2O\text{Na}_2\text{O}) is reacted completely with distilled water to prepare 200 cm3200\text{ cm}^3 of stock solution. If 50 cm350\text{ cm}^3 of this stock solution is diluted with distilled water to a final volume of 500 cm3500\text{ cm}^3 at 25∘C25^\circ\text{C}, what is the pH of the resulting diluted solution? [Na=23,O=16,H=1][\text{Na} = 23, \text{O} = 16, \text{H} = 1]

Answer: 12

Answer

The pH of the diluted solution is 12.0.
Dissolving 0.62 g0.62\text{ g} (0.01 mol0.01\text{ mol}) of Na2O\text{Na}_2\text{O} produces 0.02 mol0.02\text{ mol} of OH−\text{OH}^- ions in 200 cm3200\text{ cm}^3, giving a stock concentration of 0.1 mol dm−30.1\text{ mol dm}^{-3}. Diluting 50 cm350\text{ cm}^3 of this stock solution to 500 cm3500\text{ cm}^3 reduces the [OH−][\text{OH}^-] tenfold to 0.01 mol dm−30.01\text{ mol dm}^{-3}. Taking the negative logarithm gives pOH=2.0\text{pOH} = 2.0, which corresponds to a pH\text{pH} of 14.0−2.0=12.014.0 - 2.0 = 12.0.

Step-by-Step Solution

1
Calculate the molar mass of sodium oxide (Na2O\text{Na}_2\text{O}) and determine the amount of moles dissolved.
Molar mass of Na2O=(2×23)+16=62 g mol−1\text{Molar mass of Na}_2\text{O} = (2 \times 23) + 16 = 62\text{ g mol}^{-1}. Moles of Na2O=0.62 g62 g mol−1=0.01 mol\text{Na}_2\text{O} = \frac{0.62\text{ g}}{62\text{ g mol}^{-1}} = 0.01\text{ mol}.
Converting mass to moles is required to apply chemical stoichiometry.
2
Determine the moles of hydroxide ions (OH−\text{OH}^-) formed upon complete reaction with water.
The balanced equation is Na2O+H2O→2NaOH→2Na++2OH−\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH} \rightarrow 2\text{Na}^+ + 2\text{OH}^-. Therefore, 0.01 mol0.01\text{ mol} of Na2O\text{Na}_2\text{O} yields 0.02 mol0.02\text{ mol} of OH−\text{OH}^-.
Sodium oxide is a basic oxide that reacts with water in a 1:2 mole ratio to yield hydroxide ions.
3
Calculate the hydroxide ion concentration in the 200 cm3200\text{ cm}^3 (0.2 dm30.2\text{ dm}^3) stock solution.
[OH−]stock=0.02 mol0.2 dm3=0.1 mol dm−3[\text{OH}^-]_{\text{stock}} = \frac{0.02\text{ mol}}{0.2\text{ dm}^3} = 0.1\text{ mol dm}^{-3}.
Molarity is defined as moles of solute per cubic decimeter of solution.
4
Apply the dilution formula C1V1=C2V2C_1 V_1 = C_2 V_2 to find the hydroxide ion concentration after dilution.
[OH−]diluted=0.1 mol dm−3×50 cm3500 cm3=0.01 mol dm−3=1.0×10−2 mol dm−3[\text{OH}^-]_{\text{diluted}} = \frac{0.1\text{ mol dm}^{-3} \times 50\text{ cm}^3}{500\text{ cm}^3} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}.
Diluting 50 cm350\text{ cm}^3 to 500 cm3500\text{ cm}^3 decreases the concentration by a factor of 10.
5
Calculate the pOH and subsequently the pH of the diluted solution.
pOH=−log⁡10(1.0×10−2)=2.0\text{pOH} = -\log_{10}(1.0 \times 10^{-2}) = 2.0. Using pH+pOH=14.0\text{pH} + \text{pOH} = 14.0, pH=14.0−2.0=12.0\text{pH} = 14.0 - 2.0 = 12.0.
The logarithmic scale defines pOH=−log⁡10[OH−]\text{pOH} = -\log_{10}[\text{OH}^-] and at 25∘C25^\circ\text{C}, pH+pOH=14\text{pH} + \text{pOH} = 14.

Key Concept

Stoichiometric reaction of basic oxides with water combined with dilution calculations to determine solution pH and pOH.
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Topics

pH and pOH Scale and Calculations

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