Question

An acetic acid/ sodium acetate buffer solution similar to the one you made in the lab was prepared using the following components: 3.46 g of NaC2H3O2∙3H2O (FW. 136 g/mol) 9.0 mL of 3.0 M HC2H3O2 55.0 mL of water If you take half of this solution and add 2 mL of 1.00 M HCl to it, then what is the pH of this new solution?

Answer 1

Find out the moles of NaC₂H₃O₂.3H₂O in the original buffer solution = (mass of NaC₂H₃O₂.3H₂O)/(FW of NaC₂H₃O₂.3H₂O) = (3.46 g)/(136 g/mol) = 0.02544 mole.

We take half of the original buffer; therefore, moles of NaC₂H₃O₂.3H₂O transferred = ½*(0.02544 mole) = 0.01272 mole.

Moles of HC₂H₃O₂ transferred = (volume of HC₂H₃O₂ transferred in L)*(concentration of HC₂H₃O₂) = (9.0 mL)*(1 L/1000 mL)*(3.0 mol/L) = 0.027 mole (1 M = 1 mol/L).

Total volume of solution = (9.0 + 55.0) mL = 64.0 mL.

Concentration of HC₂H₃O₂ in the original buffer = (moles of HC₂H₃O₂)/(volume of solution in L) = (0.027 mole)/[(64.0 mL)*(1 L/1000 mL)] = 0.421875 mol/L.

Half of the volume of the buffer was transferred; therefore, volume of solution transferred = 32.0 mL. Therefore, moles of HC₂H₃O₂ transferred = (32.0 mL)*(1 L/1000 mL)*(0.421875 mol/L) = 0.0135 mole.

HCl reacts with NaC₂H₃O₂.3H₂O as below.

HCl + NaC₂H₃O₂.3H₂O --------> NaCl + HC₂H₃O₂ + 3 H₂O

As per the stoichiometric equation above,

1 mole HCl = 1 mole NaC₂H₃O₂.3H₂O = 1 mole HC₂H₃O₂

Moles of HCl added = (2.0 mL)*(1 L/1000 mL)*(1.00 mol/L) = 0.002 mole.

Moles of NaC₂H₃O₂.3H₂O neutralized = 0.002 mole; moles of NaC₂H₃O₂ retained = (0.01272 – 0.002) mole = 0.01072 mole.

Moles of HC₂H₃O₂ formed = 0.002 mole; therefore, total number of moles of HC₂H₃O₂ in the system = (0.002 + 0.0135) mole = 0.0155 mole.

Total volume of the solution = (32.0 + 2.0) mL = 34.0 mL = (34.0 mL)*(1 L/1000 mL) = 0.0.34 L.

The equilibrium concentrations are as below:

[NaC₂H₃O₂.3H₂O] = (0.01072/0.034) mol/L

[HC₂H₃O₂] = (0.0155/0.034) mol/L

pK_a of HC₂H₃O₂ = 4.75.

Use the Henderson-Hasslebach equation.

pH = pK_a + log [NaC₂H₃O₂.3H₂O]/[HC₂H₃O₂] = 4.75 + log [(0.01072/0.034) mol/L]/[(0.0155/0.034) mol/L] = 4.75 + log (0.01072/0.0155) = 4.75 + log (0.6916) = 4.75 + (-0.160) = 4.59 (ans)