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Difficulty: HardpH and pOH Scale and Calculations

A student dilutes 100 cm3100\text{ cm}^3 of a 0.05 mol dm−30.05\text{ mol dm}^{-3} nitric acid (HNO3\text{HNO}_3) solution with distilled water to achieve a total volume of 500 cm3500\text{ cm}^3 at 25∘C25^\circ\text{C}. What is the pOH of the resulting diluted solution?

  1. A
    2.02.0
  2. 12.012.0Answer
  3. C
    1.31.3
  4. D
    12.712.7

Answer

The pOH of the resulting diluted solution is 12.012.0.
Diluting 100 cm3100\text{ cm}^3 of 0.05 mol dm−3 HNO30.05\text{ mol dm}^{-3}\text{ HNO}_3 to 500 cm3500\text{ cm}^3 decreases the hydrogen ion concentration to 0.01 mol dm−30.01\text{ mol dm}^{-3}. The pH of this solution is −log⁡10(0.01)=2.0-\log_{10}(0.01) = 2.0. Since pH+pOH=14\text{pH} + \text{pOH} = 14 at 25∘C25^\circ\text{C}, the pOH is 14−2.0=12.014 - 2.0 = 12.0.

Step-by-Step Solution

1
Calculate the hydrogen ion concentration of the diluted solution using the dilution formula C1V1=C2V2C_1 V_1 = C_2 V_2.
C2=0.05 mol dm−3×100 cm3500 cm3=0.01 mol dm−3=1.0×10−2 mol dm−3C_2 = \frac{0.05\text{ mol dm}^{-3} \times 100\text{ cm}^3}{500\text{ cm}^3} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}.
Nitric acid is a strong monoprotic acid that fully dissociates in water, so [H+]=C2[\text{H}^+] = C_2.
2
Determine the pH of the diluted solution.
pH=−log⁡10[H+]=−log⁡10(1.0×10−2)=2.0\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-2}) = 2.0.
The negative logarithm of the hydrogen ion concentration defines the pH.
3
Calculate the pOH using the water ion product relationship at 25∘C25^\circ\text{C}.
pOH=14−pH=14−2.0=12.0\text{pOH} = 14 - \text{pH} = 14 - 2.0 = 12.0.
The sum of pH and pOH equals 1414 at standard room temperature.

Key Concept

Dilution effect on hydrogen ion concentration and the relation pH+pOH=14\text{pH} + \text{pOH} = 14
Estimated Time:1m 30s
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Topics

pH and pOH Scale and Calculations

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