What is the pH of the solution formed when 12.50 mL of 1.05 M KOH is added to 50.0 mL of 0.225 M HBr?

A. 0.65
B. 1.52
C. 12.48
D. 13.35

Answer: D

Explanation:

When 12.50 mL of 1.05 M KOH is added to 50.0 mL of 0.225 M HBr, the resulting solution has a pH of 13.35.

Here’s how to calculate it:

First, we need to determine the number of moles of KOH and HBr in the solution:

moles of KOH = (12.50 mL) * (1.05 mol/L) * (1 L/1000 mL) = 0.013125 mol moles of HBr = (50.0 mL) * (0.225 mol/L) * (1 L/1000 mL) = 0.01125 mol

KOH is a strong base and HBr is a strong acid, so they will react completely to form water and a salt (KBr):

KOH + HBr -> KBr + H2O

The number of moles of KOH is greater than the number of moles of HBr, so there will be an excess of KOH in the solution after the reaction is complete:

moles of excess KOH = moles of KOH - moles of HBr = 0.013125 mol - 0.01125 mol = 0.001875 mol

The total volume of the solution is the sum of the volumes of KOH and HBr:

total volume = 12.50 mL + 50.0 mL = 62.5 mL

The concentration of excess OH- ions in the solution is:

[OH-] = moles of excess KOH / total volume = 0.001875 mol / (62.5 mL * (1 L/1000 mL)) = 0.03 M

The pOH of the solution can be calculated using the formula pOH = -log[OH-]:

pOH = -log(0.03) = 1.52

The pH can be calculated using the formula pH + pOH = 14:

pH = 14 - pOH = 14 - 1.52 = 13.35

So the correct answer is D. 13.35.

report of collage model of photosynthesis process​