Answer:
a) pH = 3.54
b) pH = 3.27
c) pH = 2.71
d) pH = 2.25
e) pH = 2.04
Step-by-step explanation:
Step 1: Data given
Volume of a 0.100 M pyridine solution = 25.0 mL = 0.025 L
Kb for pyridine is 1.7 * 10^-9
Molarity of HCl = 0.100 M
Step 2: The balanced equation
C5H5N(aq) + HCl (aq) → C5H5NH+ +Cl-(aq)
Step 3: Calculate moles
Moles pyridine = Molarity pyridine * volume
Moles pyridine = 0.100 M * 0.025 L
Moles pyridine = 0.0025 moles
Moles HCl = 0.100 M * 0.0245 L
Moles HCl = 0.00245 moles
Step 4: Calculate the limiting reactant
For 1 mol pyridine we need 1 mol HCl to produce 1 mol C5H5NH+
HCl is the limiting reactant. It will completely be consumed (0.00245 moles). Pyridine is in excess. There will react 0.00245 moles. There will remain 0.0025 - 0.00245 = 0.00005 moles
There will be produce 0.00245 moles C5H5NH+
Step 5: Calculate the molarity
Molarity = moles / volume
Molarity C5H5N = 0.00005 moles / 0.0495 L
Molarity C5H5N = 0.00101 M
Molarity C5H5NH+ = 0.00245 moles / 0.0495 L
Molarity C5H5NH+ = 0.0495 M
Step 6: Calculate pOH
pOH = pKb + log ([B+]/[BOH])
pOH = 8.77 + log (0.0495 / 0.00101)
pOH = 10.46
Step 7: Calculate pH
pH = 14 - 10.46 = 3.54
When adding 25.0 mL HCl
Moles pyridine = Molarity pyridine * volume
Moles pyridine = 0.100 M * 0.025 L
Moles pyridine = 0.0025 moles
Moles HCl = 0.100 M * 0.025 L
Moles HCl = 0.0025 moles
Step 4: Calculate the limiting reactant
For 1 mol pyridine we need 1 mol HCl to produce 1 mol C5H5NH+
Both HCl and pyridine will completely be consumed.
The total volume now is 25.0 mL + 25.0 mL = 50.0 so the concentration of the C5H5NH+ to start is 0.0025 moles / 0.05 L = 0.05 M
C5H5NH+ + H2O ==> C5H5N + H3O^+
The concentration of C5H5NH+ will be 0.05 - X M
The concentration of C5H5N and H2O will be X
Ka = K/ Kb
Ka = X²/ (0.05 -X)
X = [H3O+] = 0.000542
pH = -log (0.000542)
pH = 3.27
When adding 26.0 mL HCl
Moles pyridine = Molarity pyridine * volume
Moles pyridine = 0.100 M * 0.025 L
Moles pyridine = 0.0025 moles
Moles HCl = 0.100 M * 0.026 L
Moles HCl = 0.0026 moles
Step 4: Calculate the limiting reactant
For 1 mol pyridine we need 1 mol HCl to produce 1 mol C5H5NH+
Pyridine is the limiting reactant. All of it will be consumed.
There will remain 0.0001 mol HCl
Step 5: Calculate pH of HCl
pH = -log [H+]
pH = -log (0.0001 / 0.051 L)
pH = -log ( 0.00196)
pH = 2.71
When adding 28.0 mL HCl
Moles pyridine = Molarity pyridine * volume
Moles pyridine = 0.100 M * 0.025 L
Moles pyridine = 0.0025 moles
Moles HCl = 0.100 M * 0.028 L
Moles HCl = 0.0028 moles
Step 4: Calculate the limiting reactant
For 1 mol pyridine we need 1 mol HCl to produce 1 mol C5H5NH+
Pyridine is the limiting reactant. All of it will be consumed.
There will remain 0.0003 mol HCl
Step 5: Calculate pH of HCl
pH = -log [H+]
pH = -log (0.0003 / 0.053 L)
pH = -log ( 0.00566)
pH = 2.25
When adding 30.0 mL HCl
Moles pyridine = Molarity pyridine * volume
Moles pyridine = 0.100 M * 0.025 L
Moles pyridine = 0.0025 moles
Moles HCl = 0.100 M * 0.030 L
Moles HCl = 0.0030 moles
Step 4: Calculate the limiting reactant
For 1 mol pyridine we need 1 mol HCl to produce 1 mol C5H5NH+
Pyridine is the limiting reactant. All of it will be consumed.
There will remain 0.0005 mol HCl
Step 5: Calculate pH of HCl
pH = -log [H+]
pH = -log (0.0005 / 0.055 L)
pH = -log ( 0.00909)
pH = 2.04