Final answer:
The pH of the mixed solution of 150.0 mL of 0.10 M KOH with 50.0 mL of 0.20 M HBr is determined to be approximately 12.4, as KOH is in excess and the resulting solution is basic.
Step-by-step explanation:
The student's question relates to the calculation of the pH of a solution obtained by mixing 150.0 mL of 0.10 M KOH with 50.0 mL of 0.20 M HBr. To find the pH of this solution, one must first determine if the reaction between KOH and HBr goes to completion, and then calculate the resulting concentration of the excess reactant.
KOH is a strong base and HBr is a strong acid. When these two solutions are combined, they react in a 1:1 molar ratio to produce water and KBr. Hence, for every mole of HBr, one mole of KOH is required:
KOH(aq) + HBr(aq) → KBr(aq) + H2O(l)
We can determine the moles of KOH and HBr initially present:
- Moles of KOH = 0.15 L × 0.10 M = 0.015 mol
- Moles of HBr = 0.05 L × 0.20 M = 0.010 mol
Since KOH is present in excess (0.015 mol - 0.010 mol = 0.005 mol remaining), the pH of the final solution will be basic. To calculate the concentration of the excess KOH, we have:
- Total volume of the mixture = 150.0 mL + 50.0 mL = 200.0 mL = 0.2 L
- Molarity of excess KOH = 0.005 mol / 0.2L = 0.025M
The pOH of the solution can be calculated using the formula pOH = -log[OH-], and from there, the pH can be found since pH + pOH = 14.
pOH = -log[0.025] = 1.60
Therefore, pH = 14 - pOH = 14 - 1.60 = 12.40
The pH of the solution is closest to 12.4.