Question

Consider the titration of a 37.0 ml sample of 0.175 M HBr with 0.200 M KOH. Determine each of the following:

a) Initial moles of HBr
b) Final moles of KOH
c) Volume of KOH needed for neutralization
d) pH of the resulting solution

Answer 1

The initial moles of HBr are 0.006475 mol, the final moles of KOH are also 0.006475 mol, the volume of KOH needed for neutralization is 32.375 mL, and the pH of the resulting solution at the equivalence point is 7.00.

Step-by-step explanation:

To solve the titration problem, we must apply the concept of stoichiometry and remember that HBr is a strong acid and KOH is a strong base, which will react in a 1:1 molar ratio:

• Initial moles of HBr: To find the initial moles of HBr, multiply its volume by its molarity:
• (0.037 L) × (0.175 M) = 0.006475 mol.
• Final moles of KOH: Since we are looking for the moles at the equivalence point, the final moles of KOH will be equal to the initial moles of HBr, which is 0.006475 mol.
• Volume of KOH needed for neutralization: To find the volume of KOH needed, divide the final moles of KOH by its molarity:
• (0.006475 mol) / (0.200 M) = 0.032375 L, which is 32.375 mL.
• pH of the resulting solution: Since the reaction between HBr and KOH is a complete neutralization and no excess reactants are present at the equivalence point, the resulting solution will be neutral, meaning the pH will be 7.00.

For calculations beyond the equivalence point, where the solution contains excess OH-, the concentration of OH- should be calculated similar to the remaining KOH after neutralization and then converted to pH.