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Buffers

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Mapa Mental sobre Buffers, criado por wg.fox em 10-04-2014.
wg.fox
Mapa Mental por wg.fox, atualizado more than 1 year ago
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Criado por wg.fox quase 12 anos atrás
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Resumo de Recurso

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

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