Calculating acetate buffer's pH and the amount of chemicals required to make it.

Question

1. How many milliliters of 0.1M CH3COOH(pKa=4.75) and 0.1M CH3COONa does it take to make 400 milliliters of 0.1M pH 4.5 acetate buffer?
2. What will be buffer solution's pH, when 10mL of 1M NaOH is added?
3. What would be the pH of the solution, if 10mL 1M NaOH is added to 400mL of distilled water?

Answer 1

You can use the Henderson Hasselbalch equation which is pH = pKa + log [salt]/[acid]

4.5 = 4.75 + log [salt]/[acid]

log [salt]/[acid] = -0.25

[salt]/[acid] = 0.562 

Since both reagents are 0.1 M it is simply

0.562 x 400 ml = 224.8 mls of CH3COOH + 175.2 ml of CH3COONa = 400 mls of pH 4.5 buffer.

 

2)  NaOH + CH3COOH ==> CH3COONa + H2O

moles CH3COOH initially present = 0.2248 L x 0.1 mol/L = 0.02248 moles

moles NaOH added = 0.01 L x 1.0 mol/L = 0.01 moles

moles CH3COONa formed = 0.01 moles

Final moles CH3COONa = 0.01 moles + 0.1752 L x 0.1 mol/L = 0.01 mol + 0.01752 = 0.02752 moles CH3COONa

Final moles CH3COOH = 0.02248 mol - 0.01 = 0.01248 moles CH3COOH

Final volume = 400 mls + 10 mls = 410 mls = 0.410 L

Final [salt] = 0.02752 mol/0.410 L = 0.0671 M

Final [acid] = 0.01248 mol/0.410 L = 0.0304 M

pH = 4.75 + log 0.0671/0.0394 = 4.75 + 0.231 = 4.98

 

3)  (10 ml)(1 M NaOH) = (410 ml)(x M NaOH) and x = 0.0244 M NaOH

pOH = - log 0.0244 = 1.61

pH = 14 - 1.61 = 12.4