Final answer:
a. Before the addition of HBr, the solution is basic with a pH of approximately 13.301. b. At the point of neutralization, the pH is approximately 13.177. c. After titrating 1 mL past neutralization, the pH is approximately 13.02.
Step-by-step explanation:
a. Before the addition of HBr, the solution contains 0.1 M Ca(OH)2 which dissociates into Ca2+ and 2OH-. Since Ca(OH)2 is a strong base, it completely ionizes, resulting in a solution that is basic.
The concentration of OH- can be calculated as follows:
0.1 M x 2 = 0.2 M OH-
Using the formula for the pH of a basic solution, we have:
pOH = -log([OH-]) = -log(0.2) ≈ 0.699
pH = 14 - pOH = 14 - 0.699 ≈ 13.301
b. The point of neutralization occurs when the moles of HBr added is equal to the moles of Ca(OH)2 in the solution.
Using the equation n = CV, where n is the number of moles, C is the concentration, and V is the volume, we can calculate the moles of Ca(OH)2:
n(Ca(OH)2) = 0.1 M x 50 mL = 0.005 moles
To reach the point of neutralization, we need to have added an equal number of moles of HBr, which is 0.005 moles. The volume of HBr needed can be calculated using the formula V = n/C:
V(HBr) = 0.005 moles / 0.15 M = 0.0333 L = 33.3 mL
The pH at this point can be calculated using the same method as in part (a), substituting the new concentration of OH- and solving for pH:
pOH = -log([OH-]) = -log(0.15) ≈ 0.823
pH = 14 - pOH = 14 - 0.823 ≈ 13.177
c. To determine the pH after titrating 1 mL past neutralization, we need to calculate the moles of HBr added:
n(HBr) = 0.15 M x 0.001 L = 0.00015 moles
The moles of Ca(OH)2 remaining can be calculated by subtracting the moles of HBr added from the initial moles of Ca(OH)2:
n(Ca(OH)2) remaining = 0.005 moles - 0.00015 moles = 0.00485 moles
The volume of the solution after adding 1 mL of HBr can be calculated:
V(solution) = 50 mL + 1 mL = 51 mL = 0.051 L
Using the equation n = CV with the remaining moles of Ca(OH)2 and the volume of the solution, we can calculate the new concentration of OH-:
[OH-] = n(Ca(OH)2) remaining / V(solution) = 0.00485 moles / 0.051 L = 0.095 M
The pH at this point can be calculated using the same method as before:
pOH = -log([OH-]) = -log(0.095) ≈ 0.98
pH = 14 - pOH = 14 - 0.98 ≈ 13.02