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For pure water at 40 °C, pH = 6.67 - AQA - A-Level Chemistry - Question 9 - 2019 - Paper 1

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For pure water at 40 °C, pH = 6.67. A student thought that the water was acidic. Explain why the student was incorrect. Determine the value of K_w at this tempera... show full transcript

Worked Solution & Example Answer:For pure water at 40 °C, pH = 6.67 - AQA - A-Level Chemistry - Question 9 - 2019 - Paper 1

Step 1

Explain why the student was incorrect.

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Answer

The student is incorrect because the pH of pure water at any temperature indicates neutrality at a pH of 7. A pH of 6.67 suggests that water is slightly acidic compared to pure water’s pH of 7. However, since pH can vary with temperature, it indicates that the concentration of hydrogen ions

([H⁺]) is not purely representative of acidity in comparison to the neutral state at that temperature.

Step 2

Determine the value of K_w at this temperature.

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Answer

To calculate the value of Kₓ at 40 °C, the following relationship is used: Kw=[H+][OH]K_w = [H^+][OH^-]

At pH 6.67, the concentration of hydrogen ions is: [H+]=106.67extmoldm3[H^+] = 10^{-6.67} ext{ mol dm}^{-3} Calculating this gives: [H+]approx2.13imes107extmoldm3[H^+] \\approx 2.13 imes 10^{-7} ext{ mol dm}^{-3}

Since, at equilibrium in pure water, [H+]=[OH][H^+] = [OH^-], we have: [OH]=2.13imes107extmoldm3[OH^-] = 2.13 imes 10^{-7} ext{ mol dm}^{-3}

Substituting these values into the equation for K_w: Kw=[H+][OH]=(2.13imes107)(2.13imes107)=4.53imes1014extmol2extdm6K_w = [H^+][OH^-] = (2.13 imes 10^{-7})(2.13 imes 10^{-7}) = 4.53 imes 10^{-14} ext{ mol}^2 ext{ dm}^{-6}.

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