1. Aqueous Equilibria, Part One AP Chemistry

2. large K a : Weak Acids -- most acids are weak (i.e., only partially ionized) -- For a weak acid HX... HX(aq) H + (aq) + X – (aq) -- acid-dissociation constant K a = [H + ] [X – ] [HX] small K a : “strong” weak acid “weak” weak acid The % of a weak acid that is ionized is given by the equation: [H + ] at eq. x 100% ionization = [acid] orig.

3. Formic acid is the simplest carboxylic acid. It is present in bee and ant stings. For organic acids (containing only C, H, and O), the “donated” H was connected to... O, NOT to C. First isolated from apple juice, malic acid is a dicarboxylic acid. Because of its extreme sour taste, it is used as a food additive, notably in Mega Warheads ®.

4. A 0.020 M niacin solution has pH 3.26. N C O O H niacin (a) What % of the acid is ionized? (b) What is K a ? a) [H + ] at eq. x 100% ion. = [acid] orig. 10 –3.26. x 100% ion. = 0.020 = 2.7% ionized Niacin (i.e., vitamin B 3 ) is a water-soluble vitamin and an essential nutrient. It is found naturally in such foods as meat, wheat germ, dairy products, and yeast. A niacin deficiency leads to pellagra, a condition whose symptoms include skin lesions, mental confusion, and weakness.

5. (b) What is K a ? niacin(aq) H + (aq) + conj. base – (aq) init. [ ] 0.02000  [ ] – x+ x at eq. [ ] 0.020 – xxx But, from given info, x = [H + ] = 10 –3.26 = 5.5 x 10 –4 M, so “at eq.” line becomes… at eq. [ ]0.019455.5 x 10 –4 Thus, K a = (5.5 x 10 –4 ) 2 0.01945 = 1.6 x 10 –5

6. If K a for niacin is 1.6 x 10 –5, find the pH of a 0.010 M niacin solution. niacin(aq) H + (aq) + conj. base – (aq) init. [ ] 0.01000  [ ] – x+ x at eq. [ ] 0.010 – xxx 1.6 x 10 –5 = x2x2 0.010 – x K a = P R Could use quadratic equation to solve for… x = [H + ] =3.92 x 10 –4 MpH = 3.41 Use a shortcut IF the power of K is –4, –5, –6, etc.

7. Using the shortcut method for the previous problem… 0.010 – x 1.6 x 10 –5 = x2x2 x = [H + ] =4.00 x 10 –4 M pH = 3.40 0.010 1.6 x 10 –5 = x2x2 niacin(aq) H + (aq) + conj. base – (aq) init. [ ]0.01000  [ ] – x+ x at eq. [ ]0.010 – xxx

8. % ionization of a weak acid at a given temperature... Sometimes, the less concentrated the substance, the more active is the amount that IS present. In the same way, the less concentrated the weak acid, the greater the % ionization. % ionization / as acid [ ] /

9. Calculate the % of HF molecules ionized in a 0.10 M HF solution. (K a = 6.8 x 10 –4 ) 0.1000 – x+ x HF(aq) H + (aq) + F – (aq) init. [ ]  [ ] at eq. [ ] 0.10 – xxx 6.8 x 10 –4 = x2x2 x = [H + ] = 8.25 x 10 –3 M [acid] orig. [H + ] at eq. x 100% ion.= 0.10 8.25 x 10 –3 = 8.3% (Without shortcut, % ion. is 7.9%.) X

10. Calculate the % of HF molecules ionized in a 0.010 M HF solution. (K a = 6.8 x 10 –4 ) HF(aq) H + (aq) + F – (aq) at eq. [ ]0.010 – xxx 6.8 x 10 –4 = x2x2  x = [H + ] = 2.61 x 10 –3 M [acid] orig. [H + ] at eq. x 100% ion.= 0.010 2.61 x 10 –3 = 26% (Without shortcut, % ion. is 23%.) X