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Weak Acids and Bases

Weak Acids and Bases. 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]

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Weak Acids and Bases

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  1. Weak Acids and Bases Only a portion of the weak acids and bases will break apart when in H2O 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

  2. 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 0.0 M -x X X 0.1 - 1.778 x 10-5 1.778 x 10-5 1.778 x 10-5 1 x 10-4.75 The subtracted amount is so small, we can ignore it!!

  3. 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%

  4. Example – Calculate the Ka of a 0.50 M solution of formic acid (HCHO2) with a pH of 2.02. Also determine the percent ionization in the acid. HCHO2 H+ + CHO2- 0.5 M 0.0 M 0.0 M -x X X 0.5 – 9.55 x 10-3 9.55 x 10-3 1 x 10-2.02 9.55 x 10-3 Again, the subtracted amount is so small, we can ignore it!! Ka = [H+][CHO2] [HCHO2] Ka = (9.55 x 10-3)2 0.5 Ka = 1.8 x 10-4 Percent ionized = [H+] original [HCHO2] 9.55 x 10-2 0.5 1.91%

  5. 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 0.0 M -x X X x 0.1 - x 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

  6. When we finish the problem, we should check to make sure that the ionization is less than 5%. Ka = X2 0.1 5.3 x 10-5 = X2 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 10-3x 100 0.1 2.30 % We’re safe!

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