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Acids Lesson 11 Weak Acids pH Calculations

Acids Lesson 11 Weak Acids pH Calculations. 1. Calculate the pH of 0.45 M HF. You need an ICE chart for weak acids or bases! HF ⇄ H + + F - I 0.45 M 0 0 C x x x E 0.45 - x x x Ka = [ H + ][ F - ] = 3.5 x 10 -4 [HF] Ka = x 2 = 3.5 x 10 -4 0.45 - x.

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Acids Lesson 11 Weak Acids pH Calculations

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  1. Acids Lesson 11 Weak Acids pH Calculations

  2. 1. Calculate the pH of 0.45 M HF. You need an ICE chart for weak acids or bases! HF ⇄H+ + F- I 0.45 M 0 0 C x x x E 0.45 - x x x Ka = [H+][F-] = 3.5 x 10-4 [HF] Ka = x2 = 3.5 x 10-4 0.45 - x Because the Ka is small we can assume that the x is also small This means we can make the approximation that 0.45 - x = 0.45 In other words 0.45 - 0.001 = 0.45, where 0.001 is a small number

  3. x2 = 3.5 x 10-4 0.45 x = [H+] = 0.01255 M pH = -Log[0.01255] pH = 1.90 2 sig figs due to molarity and Ka

  4. 2. Calculate the pH of 0.60 M H3BO3 H3BO3⇄H+ + H2BO3- I 0.60 M 0 0 C x x x E 0.60 - x x x 0 small ka x2 = 7.3 x 10-10 0.60 x = [H+] = 2.09 x 10-5 M pH = -Log[2.09 x 10-5] pH = 4.68 2 sig figs due to molarity and Ka

  5. 3. Calculate the pH of a 0.20 M diprotic acid with a Ka = 4.7 x 10-7 H2X ⇄H+ + HX- I 0.20 M 0 0 C x x x E 0.20 - x x x 0 small ka x2 = 4.7 x 10-7 0.20 x = [H+] = 3.066 x 10-4 M pH = -Log[3.066 x 10-4] pH = 3.51 2 sig figs due to molarity and Ka

  6. 4. Calculate the pH of a saturated solution of Mg(OH)2. This is a solubility equilibrium- no ICE Mg(OH)2⇄Mg2+ + 2OH- s s 2s Ksp = [Mg2+]][OH-]2= 5.6 x 10-12 [s][2s]2 = 5.6 x 10-12 4s3 = 5.6 x 10-12 s = 1.119 x 10-4 M 2s = [OH-] = 2.237 x 10-4 M pOH = 3.65 pH = 10.35

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