Question

The dibasic compound B (pKb1 5.00, pKb2 9.00) was titrated with 0.5 M HCl. The initial solution of B was 0.0500 M and had a volume of 50.0 mL. Find the pH at the following volumes of acid added Va = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL.

Answer 1

1. Now when 0 ml of acid is added

        B     +   H2O ⇌   HB⁺    +   OH^–

(0.05M-x)--------------(x)-------+----(x)

Kb1= x² / 0.05 = 1 × 10^-5

x² = 5 × 10^-7

x = [OH^-] = 7.071 × 10^-4

pOH = -log [OH^-] = -log 7.071 × 10^-4

pOH = 3.15; pH = 10.84-------------------------------------1

2. At 1 ml

                      B         +      H⁺       ⇌          HB⁺    

Initial        0.0025          0.0005                 0

Change    -0.0005       -0.0005          +0.0005

Final         0.002              0                   0.0005

Half way to equivalence point 1…… [B] = [HB⁺]

ICE tables use moles.

K uses molarity.

pOH = pKb1 = 5.00

pH = 9--------------------------------------------------------------------2

3. At 5 ml

              B         +      H⁺       ⇌          HB⁺    

Initial        0.0025          0.0025                 0

Change    -0.0025       -0.0025          +0.0025

Final              0                  0                   0.0025

Equivalence point 1. HB+ is both acid and base.

must either use systematic method or approx.

pOH = ½ (pKb1 +pKb2) = ½ (5+9) = 7

pH = 7-----------------------------------------------------------------------------3

4. At 9 ml

             HB⁺    +          H⁺       ⇌    H₂B²⁺

Initial        0.0025          0.0005                 0

Change    -0.0005       -0.0005          +0.0005

Final         0.002              0                   0.0005

At 9 ml we can start from E.P. 1, or from original.

Half way to eq. point 2. [H₂B²⁺] = [HB⁺]

pOH = pKb2 = 9.0

pH = 5.00-----------------------------------------------------------------------4

5. At 10ml

                     HB⁺     +          H⁺       ⇌    H₂B²⁺

Initial        0.0025          0.0025                 0

Change    -0.0025       -0.0025          +0.0025

Final              0                  0                   0.0025 -->--> 0.0025 mol in 0.06L = 0.0416M

Equivalence point 2. Only H₂B²⁺ ; must switch to Ka1.

New eqn.     H₂B²⁺               ⇌       HB⁺ + H⁺     

x dissoc. 0.0416 – x               x        x

Ka1 = Kw/Kb2 = 1.0 x 10^–14 / 1.0 x 10^–9 = 1.0 x 10^–5

Ka1 =1.0 x 10^–5 = x²/0.0416

x= [H⁺] = 6.45× 10^-4

pH= -log 6.45× 10^-4 = 3.190--------------------------------------------------5

6. At 11 ml

                     HB⁺      +          H⁺       ⇌    H₂B²⁺

Initial        0.0025          0.0055                 0

Change    -0.0055         -0.0055          +0.0055

Final         0.003                  0                   0.0025

After reacting with HB+ 0.003 mols of H+ ion will be in excess

Therefore pH of resulting solution will be pH= -log 0.003 = 2.52

1. Similarly calculated for 15,19,20and 22

pH at 15 ml= 2.3

pH at 19 ml= 2.15

pH at 20ml = 2.12

pH at 22 ml= 2.07