Question

Consider the titration of 100.0 mL of 0.400 M C5H5N by 0.200 M HCl. (Kb for C5H5N = 1.7×10-9)

Part 1 Calculate the pH after 0.0 mL of HCl added.

pH =

Part 2 Calculate the pH after 35.0 mL of HCl added.

pH =

Part 3 Calculate the pH after 75.0 mL of HCl added.

pH =

Part 4 Calculate the pH at the equivalence point.

pH =

Part 5 Calculate the pH after 300.0 mL of HCl added.

pH =

Part 6 At what volume of HCl added, does the pH = 5.23?

__mL

Answer 1

Part.1 :- . For 0.0 mL of HCl :-

ICE table of C5H5N is :

................C5H5N (aq) ..............+..............H2O (l) ----------> C5H5NH⁺ (aq) ........+................OH^- (aq)

I................0.400 M...................................................................0.0 M.......................................0.0 M

C................-X.............................................................................+X..........................................+X

E..............(0.400 - X) M ..............................................................X M........................................X M

here, X = Degree of dissociation

Expression of kb is :

Kb = [C5H5NH⁺] [OH^-] / [C5H5N]

1.7×10^-9 = X² / (0.400 - X)

Let X <<<0.400, so neglect X as compare to 0.400 , We have

X² = 1.7×10^-9× 0.400

X = (6.8 × 10^-10)^1/2

X = 2.61×10^-5

So

[OH^-] = X = 2.61×10^-5

pOH = - log [OH^-]

pOH = - log 2.61×10^-5

pOH = 4.58

pH + pOH = 14

So

pH = 14 -pOH

pH = 14 - 4.58

pH = 9.42

Part.2 :- . For 35.0 mL of HCl :-

Number of moles of HCl = Molarity x Volume = 0.200 M × 0.035 L = 0.007 moles

Number of moles of C5H5N = Molarity x Volume = 0.400 M × 0.1 L = 0.04 moles

Reaction between HCl and C5H5N is :

...............C5H5N........ +............ HCl -----------------> C5H5NH⁺................ +.......................... Cl^-

I...............0.04 moles ................0.007 moles..............0.0 moles..............................................

C.............-0.007 .......................-0.007........................+0.007 .........................................

F...............0.033 moles .............0.0 moles...................0.007 moles...............................

According to Henderson-Hasselbalch equation ,we have

pOH = pKb + log[C5H5NH⁺] / [C5H5NH]

pOH = 8.77 + log 0.007 / 0.033

pOH =8.77 + log 0.2121

pOH = 8.77 - 0.6735

pOH = 8.1

So

pH = 14 - 8.1 = 5.9

pH = 5.9

Part.3 :- . For 75.0 mL of HCl :-

Number of moles of HCl = Molarity x Volume = 0.200 M × 0.075 L = 0.015 moles

Number of moles of C5H5N = Molarity x Volume = 0.400 M × 0.1 L = 0.04 moles

Reaction between HCl and C5H5N is :

...............C5H5N........ +............ HCl -----------------> C5H5NH⁺................ +.......................... Cl^-

I...............0.04 moles ................0.015 moles..............0.0 moles..............................................

C.............-0.015 .......................-0.015........................+0.015 .........................................

F...............0.025 moles .............0.0 moles...................0.015 moles...............................

According to Henderson-Hasselbalch equation ,we have

pOH = pKb + log[C5H5NH⁺] / [C5H5NH]

pOH = 8.77 + log 0.015 / 0.025

pOH =8.77 + log 0.6

pOH = 8.77 - 0.2218

pOH = 8.55

So

pH = 14 - 8.55 = 5.45

pH = 5.45

Part.4 :- . At equivalent point :-

We know at equivalent point

M₁V₁ (C5H5N) = M₂V₂ (HCl)

V₂ = M₁V₁ / M₂

V₂ = (0.400 M ) (100.0 mL ) / 0.200 M = 200 mL = 0.2 L

i.e. Volume of HCl = 0.2 L

Number of moles of HCl = Molarity x Volume = 0.200 M × 0.2 L = 0.04 moles

Number of moles of C5H5N = Molarity x Volume = 0.400 M × 0.1 L = 0.04 moles

Reaction between HCl and C5H5N is :

...............C5H5N........ +............ HCl -----------------> C5H5NH⁺................ +.......................... Cl^-

I...............0.04 moles ................0.04 moles..............0.0 moles..............................................

C.............-0.04 .......................-0.015........................+0.04 .........................................

F...............0.0 moles .............0.0 moles...................0.04 moles...............................

Total volume = 200 mL + 100 mL = 300 mL = 0.3 L

Concentration of C5H5NH⁺ = 0.04 moles / 0.3 L = 0.1333 M

ICE table of C5H5NH⁺ is :

................C5H5NH⁺ (aq) ..............+..............H2O (l) <----------> C5H5NH (aq) ........+................H₃O⁺(aq)

I................0.1333 M...................................................................0.0 M.......................................0.0 M

C................-X.............................................................................+X..........................................+X

E..............(0.1333 - X) M ..............................................................X M........................................X M

Ka of C5H5NH⁺ = 1.0 × 10^-14 / kb = 1.0 × 10^-14 / 1.7×10^-9 = 5.88×10^-6

Ka = [C5H5NH ][H₃O⁺] / [C5H5NH⁺ ]

5.88×10^-6 = X² / (0.1333 - X)

Let X <<<0.1333, so neglect X as compare to 0.1333

X² = (5.88×10^-6 )(0.1333)

X = (0.7838×10^-6)^1/2

X = 8.85×10^-4

[H₃O⁺] = X = 8.85×10^-4

pH = -log [H₃O⁺]

pH = - log 8.85×10^-4

pH = 3.05

Part.5 :- For 300.0 mL of HCl :-

Number of moles of HCl = Molarity x Volume = 0.200 M × 0.3 L = 0.06 moles

Number of moles of C5H5N = Molarity x Volume = 0.400 M × 0.1 L = 0.04 moles

Reaction between HCl and C5H5N is :

...............C5H5N........ +............ HCl -----------------> C5H5NH⁺................ +.......................... Cl^-

I...............0.04 moles ................0.06 moles..............0.0 moles..............................................

C.............-0.04 .......................-0.04........................+0.04 .........................................

F...............0.0 moles .............0.02 moles...................0.04 moles...............................

Total volume = 0.3 L + 0.1 L = 0.4 L

Since C5H5NH⁺ is a weak acid and HCl is a strong acid, thereforre pH only depends upon strong acid i.e. HCl. We have

[H⁺] = [HCl] = 0.02 moles / 0.4 L = 0.05 M

pH = -log[H⁺]

pH = - log 0.05

pH = 1.30

Part.6 :-

we must have a buffer solution when the pH is 5.23

let's say X moles of HCl are added.

then moles of C5H5NH+ formed would = X

and moles of excess C5H5NH = 0.04 - X

pOH = 14 - 5.23 = 8.77

8.77 = 8.77 + log[X/(0.04-X)

8.77 - 8.77 = log[X/(0.04-X)

0 = log[X/(0.04-X)

10^0 = [X/(0.04-X)

1 = [X/(0.04-X)

0.04 - X = X

0.04 = 2X

X = 0.04/2 = 0.02

so, Moles of HCl added = 0.02 moles

volume of HCl added = moles/molarity = 0.02 moles / 0.200 M = 0.1L = 100 ml

Volume of HCl added = 100 mL