Question

An acetic acid/ sodium acetate buffer solution 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

-What is the total volume of the solution?

-Calculate the concentration of the [C₂H₃O₂^-] in this solution.

-Calculate the concentration of the [HC₂H₃O₂] in this solution.

-Calculate the pH of this buffer solution. The K_a for acetic acid is 1.8x10^-5.

-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?

-If you take half of this solution and add 2 mL of 1.00 M NaOH to it, then what is the pH of this new solution?

Answer 1

Total volume of the solution = 55.0+9.0 = 64.0 ml

[C2H3O2-] in the solution = (w/mwt)*(1000/V in ml)

                           = (3.46/136)*(1000/64)

                           = 0.4 M

[HC2H3O2] in the solution = M*V/total volume

                          = 3.0*9.0/64.0

              = 0.422 M

pH of acidic buffer = pka + log[C2H3O2-]/[HC2H3O2]

      pka = - logKa

          = -log(1.8*10^-5)

          = 4.74

pH = 4.74+log(0.4/0.422)

     = 4.717

in half solution

No of mol of NaC2H3O2∙3H2O present = (3.46/136)*(1/2) = 0.0127 mol

No of mol of HC2H3O2 = (3*9/1000)*(1/2) = 0.0135 mol

no of mol of HCl added = 2*1/1000 = 0.002 mol

pH of acidic buffer = pka + log[C2H3O2-]-HCl/[HC2H3O2+HCl]

                     = 4.74+log((0.0127-0.002)/(0.0135+0.002))

                     = 4.58

No of mol of NaC2H3O2∙3H2O present = (3.46/136)*(1/2) = 0.0127 mol

No of mol of HC2H3O2 = (3*9/1000)*(1/2) = 0.0135 mol

no of mol of NaOH added = 2*1/1000 = 0.002 mol

pH of acidic buffer = pka + log[C2H3O2-]+NaOH/[HC2H3O2-NaoH]

                     = 4.74+log((0.0127+0.002)/(0.0135-0.002))

                     = 4.85