physical chemistry assignment question plz help

Question

you prepare a buffer solution by  dissolving 2.00g each of benzoic acid C6H5COOH and sodium benzoate NaC6H5COO in 750.0 ml of water .What is the pH of this buffer ? Lets assume that the solutions volume is 750.0 ml

Answer 1

You can use the Henderson Hasselbach equation to determine the pH of this buffer.

pH = pKa + log{[A-]/[HA]}

 

Benzoic acid =  HA = C6H5COOH:  grams = 2.00    moles = 2.00/122 = 0.164

  Concentration [HA] = 0.164 mol/ 0.750 L = 0.0219 M

 

 

Benzoate ion =A- = C6H5COO- : grams = 2.00      moles of NaC6H5COO = 2.00/144 = 0.0139 

 Concentration [A-] = 0.0139 mol/0.75 L = 0.0185 M

 

Ka for Benzoic acid = 6.4 x 10^-5 (check tables or Internet)

 

pH = -log (6.4 x 10^-5) + log (0.0185/0.0219) = 4.19 + -0.07 = 4.12

 

Verify the calculations.