Question

a buffer solution consisting of 30ml of 1.0M HC2H3O2 + 30 ml 1.0M NaC2H3O2. 1. Write the chemical equation for buffer solution 2. Calculate ph of buffer solution and account for dilution 3. Calculate ph for buffer solution with addition of 2mL of 1.0 M HCL and account for dilution. 4. Calculate pH for buffer solution with added 2mL of 1.0 M NaOH And account for dilution. The Ka of HC2H3O2 is 1.8 x 10^-5

Answer 1

1. Equation is:

HC₂H₃O₂ (aq) +H₂O(l) H₃O⁺(aq) + C₂H₃O₂^-(aq)

2. Given that Ka = 1.8 *10^-5

pKa = -log Ka = -log ( 1.8 *10^-5) = 4.74

From Henderson Hasselbalch equation

pH = pKa + log [NaC₂H₃O₂]/[HC₂H₃O₂]

pH = 4.74 + log (1 /1) = 4.74

3. HCl is a acid which reacts with base NaC₂H₃O₂  to give HC₂H₃O₂ .

Moles of HCl added = molarity * volume in L = 1.0 M * 2/1000L = 0.002 mol

Moles of NaC₂H₃O₂ initial present = 1.0 M * 0.030 L = 0.030 mol = moles of HC₂H₃O₂

Moles of NaC₂H₃O₂ left = 0.030 - 0.002 = 0.028 mol

Moles of HC₂H₃O₂ = 0.030 + 0.002 = 0.032 mol

From Henderson Hasselbalch equation

pH = pKa + log [NaC₂H₃O₂]/[HC₂H₃O₂]

pH = 4.74 + log (0.028 / 0.032) = 4.74 - 0.06 = 4.68

4. NaOH is a base and will react with HC₂H₃O₂ to give NaC₂H₃O₂ .

Moles of NaOH added = molarity * volume in L = 1.0 M * 2/1000L = 0.002 mol

Moles of NaC₂H₃O₂ initial present = 1.0 M * 0.030 L = 0.030 mol = moles of HC₂H₃O₂

Moles of HC₂H₃O₂ left = 0.030 - 0.002 = 0.028 mol

Moles of NaC₂H₃O₂ = 0.030 + 0.002 = 0.032 mol

From Henderson Hasselbalch equation

pH = pKa + log [NaC₂H₃O₂]/[HC₂H₃O₂]

pH = 4.74 + log(0.032 / 0.028) = 4.74 + 0.06 = 4.80