A 210.0 mL buffer solution is 0.250 M in acetic acid and 0.250 M in sodium acetate.
1) What is the pH after addition of 0.0050 mol of HCL
2) What is the pH after addition of 0.0050 mol of NaOH
a)
Number of moles of CH3COOH = M*V = 0.250 M * 210 mL = 52.5
mmol
Number of moles of CH3COONa = M*V = 0.250 M * 210 mL = 52.5
mmol
Number of moles of HCl added = 0.005 mol= 5 mmol
The reaction that takes place is:
CH3COONa+ HCl
<----------------------> CH3COOH + NaCl
5 mmol of CH3COONa and HCl will react to form 1 mmol of
CH3COOH
After reaction,
number of moles of CH3COOH = 52.5 + 5 = 57.5 mmol
number of moles of CH3COONa = 52.5 - 5 = 47.5 mmol
CH3COOH and CH3COONa forms buffer
pKa of CH3COOH = 4.75
[CH3COONa ] = number of moles / total volume
= 47.5 mmol / 210 mL
= 0.226 M
[CH3COOH ] = number of moles / total volume
= 57.5 mmol / 210 mL
= 0.274 M
use:
pH = pKa + log {[CH3COONa/CH3COOH]}
= 4.75 + log
{0.226/0.274}
= 4.67
Answer: 4.67
b)
Number of moles of CH3COOH = M*V = 0.250 M * 210 mL = 52.5
mmol
Number of moles of CH3COONa = M*V = 0.250 M * 210 mL = 52.5
mmol
Number of moles of NaOH added = 0.005 mol= 5 mmol
The reaction that takes place is:
CH3COOH + NaOH
<----------------------> CH3COONa
5 mmol of CH3COOH and NaOH will react to form 1 mmol of
CH3COONa
After reaction,
number of moles of CH3COONa = 52.5 + 5 = 57.5 mmol
number of moles of CH3COOH = 52.5 - 5 = 47.5 mmol
CH3COOH and CH3COONa forms buffer
pKa of CH3COOH = 4.75
[CH3COOH] = number of moles / total volume
= 47.5 mmol / 210 mL
= 0.226 M
[CH3COONa] = number of moles / total volume
= 57.5 mmol / 210 mL
= 0.274 M
use:
pH = pKa + log {[CH3COONa/CH3COOH]}
= 4.75 + log
{0.274/0.226}
= 4.83
Answer: 4.83
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