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Comparing Acid Strength

Comparing Acid Strength. Without the Ka or Kb, we can determine the relative strengths of acids A. Binary Acids 1. In the same group Since we want them to break apart to release H + , the larger the atom, the stronger it will be HI > HBr> HCl > HF 2. In the same period

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Comparing Acid Strength

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  1. Comparing Acid Strength Without the Ka or Kb, we can determine the relative strengths of acids A. Binary Acids 1. In the same group Since we want them to break apart to release H+, the larger the atom, the stronger it will be HI > HBr> HCl > HF 2. In the same period Bond polarity is most important. More polar, breaks better Higher electroneg element = stronger acid HF > H2O > NH3 > CH4

  2. B. Oxyacids Acids that have polyatomic ions with oxygens in them. 1. With the same number of oxygens and hydrogens More electroneg the central atom, stronger acid Pulls electrons toward center more, away from H H2SO3 > H2SeO3 HClO > HBrO 2. Same central atom, same number of hydrogens, different number of oxygens More oxygens, more acidic Extra oxygens pull electrons more from the H HNO3 > HNO2 HClO < HClO2 < HClO3 < HClO4

  3. C. Carboxylic acids Organic acids contain COOH H from the OH is the atom that breaks off to create H+ Named by using the alkane name of the hydrocarbon chain, removing the “e” and adding CH3COOH As we add more electronegative atoms to the carbon chain, the atom becomes more acidic CH3COOH < CH2ClCOOH < CHCl2COOH O || X – C – OH “oic acid” “ethanoic acid”

  4. Common ion effect If we have a chemical equilibrium, then we add an ionic substance, only part of the ionic substance may affect the equilibrium. Example – in HF  H+ + F- If we add Na F Na+ F- no effectadds to products We should be able to calculate the concentration of all reactants and products after we add the common ion. Example – The Ka of HF is 6.8 x 10-4 at 25oC. What is the H+ concentration if we add enough NaF to make the solution 0.10 M HF and 0.20 M NaF? The starting [HF] is 0.1 M and the starting [F-] is 0.2 M We ignore the cation of the salt!

  5. HF  H+ + F- Due to the NaF! 0.2 M 0.1 M 0.0 M -x +x +x 0.1 x 0.2 Ka = [H+][F-] [HF] 6.8x 10-4 = 0.2 x 0.1 X = 3.4 x 10-4 M Example – What is the pH of a solution that is 0.3 M acetic acid and 0.3 M sodium acetate if the Ka for acetic acid is 1.8 x 10-5? HC2H3O2 H+ + C2H3O2- 0.3 M 0.3 M 0.0 M -x +x +x 0.3 x 0.3

  6. Ka = [H+][C2H3O2-] [HC2H3O2-] 1.8 x 10-5 = 0.3 x 0.3 X = 1.8 x 10-5 pH = -log 1.8 x 10-5 pH = 4.74 In rare cases, x may represent a large amount. If this is true, we will need to use the quadratic equation!

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