Question

Consider the titration of 100.0 mL of 0.100 M acetic acid with 0.100 M NaOH. CH3CO2H(aq) + OH-(aq) → CH3CO2-(aq) + H2O(ℓ) Ka for acetic acid = 1.8×10-5 Kb for acetate ion = 5.6×10-10 (a) What is the pH of the solution when 90.0 mL of 0.100 M NaOH has been added to 100.0 mL of 0.100 M acetic acid? (b) What is the pH at the equivalence point? (c) What is the pH after 110.0 mL of NaOH has been added?

Answer 1

Answer:

a) pH = 5.70

b) pH = 8.22

c)pH = 11.68

Explanation:

a)NaOH (aq)+CH3COOH(aq) ------> CH3COONa(aq) + H2O(aq)

1mol of NaOH react with 1mol of CH3COOH

No of mole in 90ml of 0.1M NaOH

                       = (0.1mol/1000ml)×90ml

                       = 0.009

No of mol in 100ml of CH3COOH

                     = (0.1/1000)×100

                     = 0.01

No of mol of CH3COOH after addition

                    = 0.01-0.009

                    = 0.001

Total volume = 100ml + 90ml

                     = 190ml

Final molarity of CH3COOH =( 0.0025/190) ×1000

                    = 0.00526M

Concentration of CH3COONa formed =( 0.0075/190) ×1000

                      = 0.0474M

Ka of CH3COOH = 1.8 × 10^-5

pka = -log(ka)

pKa = 4.75

Applying Henderson equation

pH = pKa + log ( [A-]/[HA])

pH = 4.75 + log ( 0.0474/0.00526)

= 5.70

b)

At equivalencepoint point ,  

No of moles of CH3COOH = 0

No of moles of CH3COO- = 0.01 mol

Total volume = 200ml

molarity of CH3COO- = 0.01/2

                     = 0.0050M

CH3COO- (aq) + H2O(l) <---------> CH3COOH(aq) + OH-(aq)

Kb = [ CH3COOH] [ OH- ] / [ CH3COO- ]

                  = 1.8 × 10^-5

[ CH3COOH ] = X

[ OH-] = X

[ CH3COO-] = 0.0050 - X

5.6 × 10^-10 = X^2/ (0.0050 - X)

we can assume , 0.0050 - X = 0.0050

5.6 × 10^-10 = X^2/0.0050

X = 1.67 × 10^6

[OH-] = 1.67 × 10^-6

pOH = 5.78

pH = 14 - pOH

pH = 14 -5.78

pH = 8.22

c) No of mol of OH from excess 10ml of NaOH = (0.1mol /1000ml)×10ml = 0.001mol

No of mol of OH- from hydrolysis of CH3COO- = (1.67×10^-6/1000)×200= 3.34×10^-7mol

Second one is negligible

So, no of mol OH- = 0.001mol

Total volume = 100ml + 110ml

                = 210ml

[OH-] =( 0.001/210)×1000

      = 0.0048M

pOH = -log[OH-]

= - log (0.0048)  

= 2.32

pH = 14 - 2.32

pH = 11.68