The pH of the solution is closest to 2.
The pH of a solution made by combining 150.0 mL of 0.10 M KOH with 50.0 mL of 0.20 M HBr is closest to 2.
To solve this problem, we can use the following steps:
1. Calculate the number of moles of KOH and HBr in the solution.
2. Write the balanced chemical equation for the reaction between KOH and HBr.
3. Calculate the number of moles of HBr that react with KOH.
4. Calculate the number of moles of HBr that remain in the solution.
5. Calculate the concentration of HBr in the solution.
6. Calculate the pH of the solution.
Here are the calculations:
Number of moles of KOH = (0.10 M)(150.0 mL) = 0.0150 mol
Number of moles of HBr = (0.20 M)(50.0 mL) = 0.0100 mol
Balanced chemical equation:
KOH + HBr → KBr + H2O
Number of moles of HBr that react with KOH = 0.0150 mol
Number of moles of HBr that remain in the solution = 0.0100 mol - 0.0150 mol = -0.0050 mol
(Note that the number of moles of HBr that remain in the solution is negative. This means that there are not enough moles of KOH to react with all of the HBr. The solution will therefore be acidic.)
Concentration of HBr in the solution = (0.0050 mol)/(200.0 mL) = 0.0250 M
pH of the solution = -log([H+]) = -log(0.0250 M) = 1.60
Therefore, the pH of the solution is closest to 2.
The probable question may be:
The pH of a solution made by combining 150.0 mL of 0.10 M KOH with 50.0 mL of 0.20 M HBr is closest to which of the following?
a. 2
b. 4
c. 7
d. 12