A 25.0 mL sample of 0.100 M pyridine (Kb for pyridine is 1.7 ✕ 10-9) is titrated with 0.100 M HCl solution. Calculate the pH after the addition of the following amounts of HCl with sig figs. a) 24.5 mL b) 25.0 mL c) 26.0 mL d) 28.0 mL e) 30.0 mL

a) pH = 3.54

b) pH =  3.27

c) pH = 2.71

d) pH = 2.25

e) pH = 2.04

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

po-215 i got it right on the quiz

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explanation: