Question

The ionization constant for water (Kw) is 9.311 × 10⁻¹⁴ at 60 °C. Calculate [H₃O⁺], [OH⁻], pH, and pOH for pure water at 60 °C.

a) [H₃O⁺] = 1.1 × 10⁻⁷ M, [OH⁻] = 9.1 × 10⁻⁸ M, pH = 6.96, pOH = 7.04
b) [H₃O⁺] = 1.1 × 10⁻⁷ M, [OH⁻] = 9.1 × 10⁻⁸ M, pH = 7.04, pOH = 6.96
c) [H₃O⁺] = 9.1 × 10⁻⁸ M, [OH⁻] = 1.1 × 10⁻⁷ M, pH = 7.04, pOH = 6.96
d) [H₃O⁺] = 9.1 × 10⁻⁸ M, [OH⁻] = 1.1 × 10⁻⁷ M, pH = 6.96, pOH = 7.04

Answer 1

At 60 °C, the ionization constant for water (Kw) is 9.311 × 10⁻¹⁴. Using this value, we can calculate the concentration of hydronium ions, hydroxide ions, pH, and pOH for pure water at this temperature. The correct values are: [H₃O⁺] = 9.1 × 10⁻⁸ M, [OH⁻] = 1.1 × 10⁻⁷ M, pH = 6.96, and pOH = 7.04.

Step-by-step explanation:

The ionization constant for water (Kw) is a measure of how much water dissociates into hydronium ions (H3O+) and hydroxide ions (OH-) at a given temperature. At 60 °C, the Kw value is 9.311 × 10⁻¹⁴.

1. To calculate the concentration of hydronium ions, [H₃O⁺], we can use the formula: [H₃O⁺] = √Kw. Plugging in the value of Kw, we get [H₃O⁺] = √(9.311 × 10⁻¹⁴) = 1.1 × 10⁻⁷ M.
2. Similarly, the concentration of hydroxide ions, [OH⁻], can be calculated using the formula: [OH⁻] = √Kw. Substituting Kw, we get [OH⁻] = √(9.311 × 10⁻¹⁴) = 9.1 × 10⁻⁸ M.
3. The pH is a measure of the acidity of a solution and is calculated using the formula: pH = -log[H₃O⁺]. Using the value of [H₃O⁺] we calculated earlier, we find pH = -log(1.1 × 10⁻⁷) = 6.96.
4. The pOH is a measure of the basicity of a solution and is calculated using the formula: pOH = -log[OH⁻]. Using the value of [OH⁻] we calculated earlier, we find pOH = -log(9.1 × 10⁻⁸) = 7.04.

Therefore, the correct option is d) [H₃O⁺] = 9.1 × 10⁻⁸ M, [OH⁻] = 1.1 × 10⁻⁷ M, pH = 6.96, pOH = 7.04.