Question

Kw = 5.48 × 10-⁻¹⁴ at 50 °C. A solution is prepared by dissolving 0.030 mol of Ba(OH)2 in 1.50 L of aqueous solution at 50 °C. What is the pH of the solution?

Answer 1

The pH of the solution at 50 °C is calculated to be 11.86 by determining the concentration of hydroxide ions produced from Ba(OH)2 and then using the ion-product constant for water at this temperature to find the hydronium ion concentration.

Step-by-step explanation:

To calculate the pH of the solution at 50 °C, we first need to determine the concentration of hydroxide ions ([OH−]) from the dissolved Ba(OH)2, which is a strong base. Ba(OH)2 fully dissociates into Ba2+ and 2OH−, so the concentration of [OH−] will be twice the concentration of Ba(OH)2 because each unit of Ba(OH)2 produces two OH− ions. The molar concentration of Ba(OH)2 is given by dividing the number of moles by the volume of the solution: 0.030 mol / 1.50 L = 0.020 M. Therefore, the [OH−] = 2 × 0.020 M = 0.040 M.

Now, using the ion-product constant for water (Kw) at 50 °C, which is 5.48 × 10−14, we can find the [H3O+] by rearranging the expression for Kw: Kw = [H3O+][OH−]. Solving for [H3O+] gives us [H3O+] = Kw / [OH−] = (5.48 × 10−14) / 0.040 M = 1.37 × 10−12 M.

The pH of the solution is then obtained by taking the negative logarithm of the [H3O+] concentration: pH = −log[H3O+] = −log(1.37 × 10−12) = 11.86. Hence, the pH of the solution at 50 °C is 11.86.