A solution containing 0.40 M B (a weak base) and 0.20 M BH+ (the salt of the weak
base) has a pH = 10.30. What is the pH after 25 mmol of HCl is added to 250 mL of this
solution?
First use the Henderson-Hasselbalch equation and solve for pKa.
pH = pKa + log (base)/(acid)
10.30 = pKa + log [0.4/0.2]
pKa = ?
mmols B initially = 0.4 x 250 = 100
mmols BH^+ initially = 0.2 x 250 = 50
..........B + H^+ ==> BH^+
I........100...0......50
add............25..........
C........-25...-25....+25
E.........75...0.......75
Using pKa from above, substitute the E line into a new HH equation and solve for pH.
Since B = BH^+ then the log term will be zero and the new pH will equal pKa.
10
To find the pH after adding HCl to the solution, we need to calculate the concentration of the solution and then use the Henderson-Hasselbalch equation.
Step 1: Calculate the initial concentrations of B and BH+.
Given:
Initial concentration of B (weak base) = 0.40 M
Initial concentration of BH+ (salt of weak base) = 0.20 M
Step 2: Calculate the volume of the solution.
Given:
Volume of the solution = 250 mL
Since we have the volume in milliliters, we need to convert it to liters:
Volume of the solution = 250 mL ÷ 1000 mL/L = 0.250 L
Step 3: Calculate the moles of B and BH+.
Moles of B = initial concentration of B × volume of the solution
Moles of B = 0.40 M × 0.250 L = 0.10 mol
Moles of BH+ = initial concentration of BH+ × volume of the solution
Moles of BH+ = 0.20 M × 0.250 L = 0.05 mol
Step 4: Calculate the concentration of B and BH+ after adding HCl.
Given:
Amount of HCl added = 25 mmol
We need to convert the amount of HCl to moles:
Amount of HCl = 25 mmol ÷ 1000 = 0.025 mol
Since HCl is a strong acid, it completely dissociates:
HCl(aq) → H+(aq) + Cl-(aq)
Therefore, the moles of H+ produced will be equal to the moles of HCl added:
Moles of H+ = 0.025 mol
Step 5: Calculate the new moles of B and BH+.
After adding HCl, some of the B and BH+ will react with H+ to form the conjugate acid of B, BH+.
Moles of B remaining = initial moles of B - moles of H+
Moles of B remaining = 0.10 mol - 0.025 mol = 0.075 mol
Moles of BH+ remaining = initial moles of BH+ - moles of H+
Moles of BH+ remaining = 0.05 mol - 0.025 mol = 0.025 mol
Step 6: Calculate the new concentrations of B and BH+.
Concentration of B remaining = moles of B remaining ÷ volume of the solution
Concentration of B remaining = 0.075 mol ÷ 0.250 L = 0.30 M
Concentration of BH+ remaining = moles of BH+ remaining ÷ volume of the solution
Concentration of BH+ remaining = 0.025 mol ÷ 0.250 L = 0.10 M
Step 7: Use the Henderson-Hasselbalch equation to calculate the pH.
pH = pKa + log([BH+]/[B])
In this case, BH+ is the conjugate acid of B, so we can use the pKa of BH+.
Given:
pKa of BH+ = 14 - pH = 14 - 10.30 = 3.70
pH = 3.70 + log(0.10 M/0.30 M)
pH = 3.70 + log(0.333)
pH ≈ 4.02
Therefore, the pH after adding 25 mmol of HCl to 250 mL of the solution is approximately 4.02.