Buffers

Buffers

1. Weak acid and one of its salt

   • Calculate pH of buffer
     • Calculate the pH of solution containing 0.1m ethanoic acid and 0.05M sodium ethanoate (Ka for ethanoic acid is 1.7x10-5)
       • Ka = [H+]*[CH3COO-]/[CH3COOH] = [H+]*(0.1)/(0.05) = 1.75x10-5
         • [H+] = 1.7x10-5 x 0.05/0.1 = 8.5x10-6. pH = -log(8.5x10-6) = 5.07
           • Adding acid and base to the buffer
             • 20cm3 of 0.5M CH3COOH with 15cm3 of 0.5M NaOH
               • Moles of NaOH added = CV/1000 = 0.5x15/1000 = 7.5x10-2
                 • Moles of CH3COOH = 0.5 x 20/1000 = 10x10-3
                   • Moles of CH3COOH remaining = (10-7.5)x10-3 = 2.5x10-3
                     • Moles of CH3COO- produced = 7.5x10-3 (Amount of CH3COOH reacted)
                       • Ka = [H+]*[CH3COO-]/[CH3COOH] = [H+]*(7.5x10-3)/(2.5x10-3) = 1.7x10-5
                         • [H+] = 1.7x10-5 * 2.5x10-3 / 7.5x10-3 = 5.67x10-6. pH = -log10(5.67x10-6)
                           • pH = 5.25
   • Adding acid to the buffer
     • Adding 1cm3 HCl to 1.0dm3 of a buffer of ethanoic acid and sodium ethanoate of concentration 0.1mol/dm3 of each (pka = 4.75)
       • CH3COOH + CH3COO- + H+. Ka = [CH3COO-]*[H+]/[CH3COOH]
         • Ka = 0.1 x [H+]/0.1 = 1.778x10-5. Og pH = -log(1.778x10-5) = 4.75 (pKa = Ka)
           • moles of HCl = CV/1000 = 1x1/1000 = 0.001 moles H+ added
             • 1.018x10-3 moles in 1.001 dm-3. -log10(1.0168x10-3) = 2.99