1)
Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {0.1/0.2}
= 4.444
Answer: 4.44
2)
mol of HCl added = 0.031M *21.2 mL = 0.6572 mmol
CH3COO- will react with H+ to form CH3COOH
Before Reaction:
mol of CH3COO- = 0.1 M *100.0 mL
mol of CH3COO- = 10 mmol
mol of CH3COOH = 0.2 M *100.0 mL
mol of CH3COOH = 20 mmol
after reaction,
mol of CH3COO- = mol present initially - mol added
mmol of CH3COO- = (10 - 0.6572) mmol
mol of CH3COO- = 9.3428 mmol
mol of CH3COOH = mol present initially + mol added
mol of CH3COOH = (20 + 0.6572) mmol
mol of CH3COOH = 20.6572 mmol
Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745
since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {9.343/20.66}
= 4.4
Answer: 4.40
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