Question

The ionization constant for water (Kw) is 2.9 × 10⁻¹⁴ at 40 °C. Calculate [H₃O⁺], [OH⁻], pH, and pOH for pure water at 40 °C.

a) [H₃O⁺] = 1.7 × 10⁻⁷ M, [OH⁻] = 5.88 × 10⁻⁸ M, pH = 7, pOH = 7
b) [H₃O⁺] = 1.7 × 10⁻⁷ M, [OH⁻] = 5.88 × 10⁻⁸ M, pH = 7.77, pOH = 6.23
c) [H₃O⁺] = 5.88 × 10⁻⁸ M, [OH⁻] = 1.7 × 10⁻⁷ M, pH = 6.23, pOH = 7.77
d) [H₃O⁺] = 5.88 × 10⁻⁸ M, [OH⁻] = 1.7 × 10⁻⁷ M, pH = 7, pOH = 7

Answer 1

In a sample of pure water at 40 °C, the concentration of hydronium ions and hydroxide ions is equal, resulting in a neutral solution. The concentration of hydronium ions and hydroxide ions can be calculated using the square root of the ionization constant for water (Kw). The pH and pOH of the solution can then be determined using the concentrations of hydronium and hydroxide ions.

Step-by-step explanation:

In a sample of pure water at 40 °C, the concentration of hydronium ions ([H₃O⁺]) and hydroxide ions ([OH⁻]) will be equal. This is because the autoionization constant of water (Kw) at 40 °C is 2.9 × 10⁻¹⁴. To find the individual concentrations of [H₃O⁺] and [OH⁻], we can take the square root of Kw.

[H₃O⁺] = [OH⁻] = √(2.9 × 10⁻¹⁴) = 1.7 × 10⁻⁷ M

The pH of a solution is a measure of the concentration of hydronium ions. It can be calculated using the formula pH = -log[H₃O⁺]. In this case, pH = -log(1.7 × 10⁻⁷) = 7.

pOH is the negative logarithm of the concentration of hydroxide ions and can be calculated using the formula pOH = -log[OH⁻]. In this case, pOH = -log(1.7 × 10⁻⁷) = 7.