What is the ph of pure water at 40.0°c if the kw at this temperature is 2.92 × 10-14?

Chemistry

1 answer:

olganol [36]3 years ago

5 0

Answer: pH = 6.77

Explanation:

1) Chemical equilibrium

2 H₂O (l) ⇄ H₃O⁺ (aq) + OH⁻ (aq)

2) Equilibrium constant, Kw

Kw = [H₃O⁺] × [OH⁻]

By stoichiometry [H₃O⁺] = [OH⁻]. Call it x

Kw = x²

x² = 2.92 × 10⁻¹⁴ M²

x = √ (2.92 × 10⁻¹⁴) = 1.709 × 10⁻⁷ M = [H₃O⁺]

3) pH

pH = - log [H₃O⁺] = - log (1.709 × 10⁻⁷) = 6.77

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The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

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Given; 3.0 ml of 1.0 m hcl.

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0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$