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Only a portion of the weak acids and bases will break apart when in H 2 O Only a fraction of the molecules will create H + or OH - Use the Keq to determine how much of the chemical breaks up HF (aq) H + (aq) + F - (aq) Keq = [H + ][F - ] [HF] This Keq = Ka Ionization constant for acids Higher Ka value = more breaking apart of the acid = more H + in water = STRONGER ACID

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2. Calculating Ka from pH If we know the pH, we can get the Ka 1. Use the equilibrium chart 2. Use pH to determine [H + ] at equilibrium 3. Determine other concentrations at equilibrium 4. Use Ka expression to calculate Ka Example If the pH of a 0.1 M HX is 4.75, what is Ka? HX H + + X - 0.1 M 0.0 M -x X X 1 x 10 -4.75 1.778 x 10 -5 0.1 - 1.778 x 10 -5 The subtracted amount is so small, we can ignore it!!

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Ka = [H + ][X - ] [HX] 3. Determining the Percent ionized acid If we need to know what percent of the acid is ionized % =. [H + ]. original concentration Percent ionized = 1.778 x 10 -5 x 100 0.1 Ka = (1.778 x 10 -5 ) 2 0.1 Ka = 3.16 x 10 -9 0.0178%

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Example – Calculate the Ka of a 0.50 M solution of formic acid (HCHO 2 ) with a pH of 2.02. Also determine the percent ionization in the acid. HCHO 2 H + + CHO 2 - 0.5 M 0.0 M -x X X 1 x 10 -2.02 9.55 x 10 -3 0.5 – 9.55 x 10 -3 Again, the subtracted amount is so small, we can ignore it!! Ka = (9.55 x 10 -3 ) 2 0.5 Ka = 1.8 x 10 -4 Ka = [H + ][CHO 2 ] [HCHO 2 ] Percent ionized = [H + ] original [HCHO 2 ] 9.55 x 10 -2 0.5 1.91%

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4. Finding pH from Ka Done much the same way 1. Set up the chart 2. Determine final concentration in terms of x 3. Us Ka to solve for x and final [ ] of all products 4. Take the –log of the final [H + ] Example – If HY has a Ka of 5.3 x 10 -5, what is the pH of a 0.1 M solution? HY H + Y - 0.1 M 0.0 M -x X X x x 0.1 - x But, in previous problems, 0.1 – x is essentially 0.1 If ionization is less than 5%, we can ignore the amount of ionized acid from the original amount

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When we finish the problem, we should check to make sure that the ionization is less than 5%. Ka = X 2 0.1 5.3 x 10 -5 = X 2 0.1 X = 2.30 x 10 -3 This means that [H + ] is 2.30 x 10 -3 M pH = -log (2.3 x 10 -3 ) pH = 2.64 Now check to make sure ionization is less than 5% % ionized = [H + ] x 100 [HY] 2.30 x 10 -3 x 100 0.1 2.30 % We’re safe!

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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria

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