1 You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH)
ID: 882101 • Letter: 1
Question
1 You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [Ka = 1.8 × 10–5]
2 Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C6H5COOH) and 0.20 M sodium benzoate (C6H5COONa), what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (Ka (C6H5COOH) = 6.5 x 10-5)
Someone please explain these to me in FINE detail, i have an exam tomorrow and need help
Explanation / Answer
1). Given :
Volume of buffer solution = 500.0 mL = 0.5000 L
[CH3COOH]= 0.30 M
[CH3COONa]= 0.20 M
Volume of NaOH = 20.0 mL ,
[NaOH]= 1.00 M
Ka of CH3COOH = 1.8E-5
Solution:
Calculation of moles of CH3COOH , NaOH and CH3COONa
Mol CH3COOH = Molarity * volume in L
= 0.30 M * 0.5000 L
= 0.15 mol
Mol CH3COONa = 0.20 mol * 0.5000 L
= 0.10 mol
Mol NaOH = 1.00 M * 0.020 L = 0.02 mol
The reaction :
CH3COOH + NaOH --- > CH3COONa + H2O
From this reaction we say that when 0.02 mol NaOH is added to buffer then 0.02 mol Acetic acid will decrease and 0.02 mol of CH3COONa will be increased.
We use Henderson-Hasselbalch equation
pH = pka + log ([CH3COO-]/ [CH3COOH] )
[CH3COO-] = [CH3COONa]
Volume is same so we just use mole ratio
log ([CH3COO-]/ [CH3COOH] ) = log (mol CH3COO- / mol CH3COOH)
pH = -log (ka ) + log (0.10 + 0.020/ 0.15 – 0.020 )
pH = -log (1.8E-5) + log (0.10 + 0.020/ 0.15 – 0.020 )
= 4.74 + (-0.03476)
=4.70
pH of the solution = 4.70
Q. 2 )
Mol of benzoic acid = 0.250 M * 0.250 L
= 0.0625 mol benzoic acid
Mol sodium benzoate
= 0.20 M * 0.250 L
= 0.050 mol
Mol HCl = 0.1000 M * 0.025 L
= 0.0025 mol
The reaction
C6H5COONa + HCl ---- > C6H5COOH + NaCl
From this reaction
Final moles of benzoic acid = 0.0625 mol + 0.0025 mol
Final moles of sodium benzoate = 0.0500 mol – 0.0025 mol
pH = 4.74 + log (0.0500 mol – 0.0025 mol / 0.0500 mol – 0.0025 mol )
pH = 4.60
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