Final answer:
The pH of a mixture formed by adding 20.0 mL of 0.015 M Ba(OH)₂ with 40.0 mL of 8.2 x 10⁻³ M KOH is 11.87.
Step-by-step explanation:
To calculate the pH of the mixture, we need to first determine the concentration of hydroxide ions in the solution. We can do this by adding the concentrations of hydroxide ions from both the Ba(OH)₂ and KOH solutions.
For the Ba(OH)₂ solution:
Concentration = 0.015 M * (20.0 mL / (20.0 mL + 40.0 mL)) = 0.005 M
For the KOH solution:
Concentration = 8.2 x 10⁻³ M * (40.0 mL / (20.0 mL + 40.0 mL)) = 2.46 x 10⁻³ M
Now we can calculate the OH⁻ concentration in the mixture:
OH⁻ concentration = 0.005 M + 2.46 x 10⁻³ M = 0.00746 M
Finally, we can calculate the pH using the formula pH = 14 - pOH. Since pOH = -log(OH⁻) and OH⁻ concentration = 0.00746 M, we can find pOH = -log(0.00746) = 2.13. Therefore, pH = 14 - 2.13 = 11.87.