1. Using the ICE chart and/or Henderson-Hasselbalch equation, calculate the pH o
ID: 926708 • Letter: 1
Question
1. Using the ICE chart and/or Henderson-Hasselbalch equation, calculate the pH of the buffer before and after each addition of .10M NaOH solution ( 0 mL, .5 mL, 1 mL, 1.5 mL, 2.0 mL, 2.5 mL). You my show all your work to receive any credit
Part B: Preparation of a buffer and measuring its pH
1. Bring 8 mL of 0.10 M acetic acid and 8 mL of 0.10 M sodium acetate in two
labeled beakers to your lab bench. Using these solutions, you will be preparing
three different buffers and measuring their pH so it is important you label your
beakers.
2. Using a 5 mL volumetric pipet, add 1.0 mL of 0.10 M acetic acid to a medium test
tube. To the same test tube, add 4.0 mL of 0.10 M sodium acetate using another 5
mL volumetric pipet. You have just prepared an acetic acid/ sodium acetate
buffer
3. Now measure the pH of your buffer using a pH meter and record this value in the
data table. Please read the instructions above to properly use the pH meter. Your
instructor will show a demonstration during their pre-lab lecture.
4. Discard the solution in the waste container.
5. In a second medium test tube, add 4.0 mL of 0.10 M acetic acid and 1.0 mL of
0.10 M sodium acetate. Measure and record the pH. Discard the solution in the
waste container.
6. In a third medium test tube, add 2.5 mL of 0.10 M acetic acid and 2.5 mL of 0.10
M sodium acetate. Measure and record the pH
Explanation / Answer
The Henderson-Hasselbalch Equation is as follows:
pH = pKa + log ([A-]/[HA]) = pKa + log ([salt form] / [acid form])
[A-] = molar concentration of a conjugate base
[HA] = molar concentration of a undissociated weak acid (M)
(a) Buffer Solution prepared in Part B, Step 6
Molarity of acetic acid = 0.1 M
2.5 ml of acetic acid = 0.1 (mol/L) * 2.5 (ml) = 0.25 mmol
Molarity of sodium acetate = 0.1 M
2.5 ml of sodium acetate = 0.1 (mol/L) * 2.5 (ml) = 0.25 mmol
Now,
pH = -log Ka + log ([salt form] / acid form])
pH = - log (1.8 * 10-5 ) + log [0..25/0..25] = 4.75
Addition of 0.5ml NaOH
Molarity of NaOH = 0.1M
0.5 ml of NaOH = 0.1 (mol/L) * 0.5 (ml) = 0.05 mmol
The NaOH added will reduce the quantity of acetic acid by reacting with it
CH3COOH + NaOH ----> CH3COONa + H2O
Final no. of moles of acetic acid = 0.25 – 0.05 = 0.2
The NaOH added will increase the quantity of sodium acetate
Final no. of moles of sodium acetate = 0.25 + 0.05 = 0.3
Now,
pH = -log Ka + log ([salt form] / acid form])
pH = - log (1.8 * 10-5 ) + log [0.3/0.2] = 4.75 + 0.176 = 4.92
Addition of 1ml NaOH
Molarity of NaOH = 0.1M
1 ml of NaOH = 0.1 (mol/L) * 1 (ml) = 0.1 mmol
The NaOH added will reduce the quantity of acetic acid by reacting with it
CH3COOH + NaOH ----> CH3COONa + H2O
Final no. of moles of acetic acid = 0.25 – 0.1 = 0.15
The NaOH added will increase the quantity of sodium acetate
Final no. of moles of sodium acetate = 0.25 + 0.1 = 0.35
Now,
pH = -log Ka + log ([salt form] / acid form])
pH = - log (1.8 * 10-5 ) + log [0.35/0.15] = 4.75 + = 5.12
Addition of 1.5ml NaOH
Molarity of NaOH = 0.1M
1.5 ml of NaOH = 0.1 (mol/L) * 1.5 (ml) = 0.15 mmol
The NaOH added will reduce the quantity of acetic acid by reacting with it
CH3COOH + NaOH -----> CH3COONa + H2O
Final no. of moles of acetic acid = 0.25 – 0.15 = 0.1
The NaOH added will increase the quantity of sodium acetate
Final no. of moles of sodium acetate = 0.25 + 0.15 = 0.4
Now,
pH = -log Ka + log ([salt form] / acid form])
pH = - log (1.8 * 10-5 ) + log [0.4/0.1] = 4.75 + = 5.35
Addition of 2ml NaOH
Molarity of NaOH = 0.1M
2 ml of NaOH = 0.1 (mol/L) * 1 (ml) = 0.2 mmol
The NaOH added will reduce the quantity of acetic acid by reacting with it
CH3COOH + NaOH ---- > CH3COONa + H2O
Final no. of moles of acetic acid = 0.25 – 0.2 = 0.05
The NaOH added will increase the quantity of sodium acetate
Final no. of moles of sodium acetate = 0.25 + 0.2 = 0.45
Now,
pH = -log Ka + log ([salt form] / acid form])
pH = - log (1.8 * 10-5 ) + log [0.45/0.05] = 4.75 + = 5.704
Addition of 2.5ml NaOH
On addition of 2.5 ml of NaOH, all acetic acid will be finished.
The NaOH added will increase the quantity of sodium acetate
Final no. of moles of sodium acetate = 0.25 + 0.25 = 0.5 mmol
Total volume = 7.5 ml
Concentration = 0.5/7.5 = 0.066 M
pKw = pKa + pKb
14.00 = 4.75 + pKb
pKb = 9.24
Kb= 10-pKb = 10-9.24 = 5.75×10-10
[OH-] = (5.75×10-10 x 0.066)0.5 = 0.619 x 10-5
P[OH] = - log [OH-] = -log[0.619 x 10-5] = 5.2
pH = 8.8
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