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Buffers

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Mindmap am Buffers, erstellt von wg.fox am 10/04/2014.
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Mindmap von wg.fox, aktualisiert more than 1 year ago
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Zusammenfassung der Ressource

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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