The ionic product of water, K_w = 2.93 × 10^{-15} mol² dm^{-6} at 10 °C
0 3 - AQA - A-Level Chemistry - Question 3 - 2017 - Paper 1
Question 3
The ionic product of water, K_w = 2.93 × 10^{-15} mol² dm^{-6} at 10 °C
0 3 . 1 Which is the correct expression for K_w?
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A K_w = [H_2O] [H^+]
B... show full transcript
Worked Solution & Example Answer:The ionic product of water, K_w = 2.93 × 10^{-15} mol² dm^{-6} at 10 °C
0 3 - AQA - A-Level Chemistry - Question 3 - 2017 - Paper 1
Step 1
Which is the correct expression for K_w?
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Answer
The correct expression for the ionic product of water, Kw, is given by option C:
Kw=[H+][OH−]
Step 2
Calculate the pH of pure water at 10 °C
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Answer
To calculate the pH of pure water, we use the formula for Kw:
ext{Thus, } pH = - ext{log}[H^+] = - ext{log}(1.18 \times 10^{-7}) \approx 6.93.$$
The final answer is:
pH of pure water at 10 °C is 6.93.
Step 3
Suggest why this pure water at 10 °C is not alkaline.
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Answer
Pure water is not alkaline because the concentration of hydroxide ions, [OH−], equals the concentration of hydrogen ions, [H+], at equilibrium. Therefore, the solution remains neutral with a pH of around 7, which is not alkaline.
Step 4
Calculate the pH of a 0.00131 mol dm^{-3} solution of calcium hydroxide at 10 °C
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Answer
For a calcium hydroxide solution:
ext{Calculating } [H^+]:\
\text{Using } K_w: \
[H^+] = \frac{K_w}{[OH^-]} = \frac{2.93 \times 10^{-15}}{0.00262} = 1.12 \times 10^{-12} mol dm^{-3}.\
pH = - ext{log}[H^+] = - ext{log}(1.12 \times 10^{-12}) \approx 11.95.$$
The final answer is:
pH of solution at 10 °C is 11.95.
Step 5
Predict whether the pH of the magnesium hydroxide solution formed at 10 °C is larger than, smaller than or the same as the pH of the calcium hydroxide solution at 10 °C.
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Answer
The pH of magnesium hydroxide solution is expected to be smaller than the pH of the calcium hydroxide solution. This is due to magnesium hydroxide's lower solubility in water compared to calcium hydroxide, leading to a lower concentration of hydroxide ions in the solution.
