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
