1. A 1.00 L solution is prepared by placing 0.10 moles of a weak acid (HA) and 0.050 moles of its conjugate base NaA. Determine the pH of the solution. The Ka of the HA is 1.4 x 10 -5. Chemistry: HA + H 2 O ↔ H 3 O + + A - I. C. E. 0.10 0 0.050 -x +x +x 0.10-x x 0.050+x Try dropping the +x and the –x terms (5 % rule). X = [H 3 O + ] = 2.8 x 10 -5 pH = -log[x] = 4.55

2. A 1.00 L solution is prepared by placing 0.10 moles of a weak base (B) and 0.050 moles of its conjugate acid BHCl. Determine the pH of the solution. The K b of the B is 1.4 x 10 -5. Chemistry: B + H 2 O ↔ OH - + BH + I. C. E. 0.10 0 0.050 -x +x +x 0.10-x x 0.050+x Try dropping the +x and the –x terms (5 % rule). X = [OH - ] = 2.8 x 10 -5 pOH = -log[x] = 4.55 and pH = 9.45

3. A solution is 0.095 M in ascorbic acid and 0.055 M in sodium hydrogen ascorbate. Determine the pH of the solution. K a1 of the H 2 C 6 H 6 O 6 is 6.8 x 10 -5, K a2 = 2.8 x10 -12. Chemistry: H 2 C 6 H 6 O 6 + H 2 O ↔ H 3 O + + H C 6 H 6 O 6 - I. C. E. 0.095 0 0.055 -x +x +x 0.095-x x 0.055+x Try dropping the +x and the –x terms (5 % rule). X = [H 3 O + ] = 1.17 x 10 -4 pH = -log[x] = 3.94

4. A weak acid, HA (K a = 1.6 x 10 -5 ), is being titrated with a strong base, KOH. Determine the pH of the solution resulting from the mixing of 30.00 mL of 0.10 M HA with 12.00 mL of 0.10 M KOH. First we must determine what is present after the initial reaction between the acid and the base. first reaction: HA(aq) + KOH(aq) → H 2 O(l) + KA(aq) + XS-HA?? Find starting moles of HA and moles of KOH. Starting moles HA: Starting moles KOH: Now, Let’s examine the resulting equilibrium. Therefore, we have 1.8 x 10 -3 moles of HA left and 1.2 x 10 -3 moles of its conjugate base (A - ) formed. We can now change these amounts to Molarity and use them in our equilibrium expression.

5. A weak acid, HA (K a = 1.6 x 10 -5 ), is being titrated with a strong base, KOH. Determine the pH of the solution resulting from the mixing of 30.00 mL of 0.10 M HA with 12.00 mL of 0.10 M KOH. first reaction: HA(aq) + KOH(aq) → H 2 O(l) + KA(aq) + XS-HA?? Find starting moles of HA and moles of KOH. Starting moles HA: Starting moles KOH: Now, Let’s examine the resulting equilibrium. Therefore, we have 1.8 x 10 -3 moles of HA left and 1.2 x 10 -3 moles of its conjugate base (A - ) formed. We can now change these amounts to Molarity and use them in our equilibrium expression. HA + H 2 O ↔ H 3 O + + A - I. C. E. 0.0429 0 0.0286 -x +x +x 0.0429-x x 0.0286+x

6. A weak acid, HA (K a = 1.6 x 10 -5 ), is being titrated with a strong base, KOH. Determine the pH of the solution resulting from the mixing of 30.00 mL of 0.10 M HA with 12.00 mL of 0.10 M KOH. Now, Let’s examine the resulting equilibrium. Therefore, we have 1.8 x 10 -3 moles of HA left and 1.2 x 10 -3 moles of its conjugate base (A - ) formed. We can now change these amounts to Molarity and use them in our equilibrium expression. HA + H 2 O ↔ H 3 O + + A - I. C. E. 0.0429 0 0.0286 -x +x +x 0.0429-x x 0.0286+x Try dropping the -x and +x terms. If the value of x comes out to less than 5% of 0.0286, dropping the term is justified. x = [H 3 O + ] = 2.4 x 10 -5 pH = -log [2.4 x 10 -5 ] = 4.62

7. A weak acid, HA (K a = 1.6 x 10 -5 ), is being titrated with a strong base, KOH. Now try: Determine the pH of the solution resulting from the mixing of 30.00 mL of 0.10 M HA with 22.00 mL of 0.10 M KOH. First we must determine what is present after the initial reaction between the acid and the base. first reaction: HA(aq) + KOH(aq) → H 2 O(l) + KA(aq) + XS-HA?? Find starting moles of HA and moles of KOH. Starting moles HA: Starting moles KOH: Now, Let’s examine the resulting equilibrium. Therefore, we have 0.8 x 10 -3 moles of HA left and 2.2 x 10 -3 moles of its conjugate base (A - ) formed. We can now change these amounts to Molarity and use them in our equilibrium expression.

8. A weak acid, HA (K a = 1.6 x 10 -5 ), is being titrated with a strong base, KOH. Determine the pH of the solution resulting from the mixing of 30.00 mL of 0.10 M HA with 12.00 mL of 0.10 M KOH. first reaction: HA(aq) + KOH(aq) → H 2 O(l) + KA(aq) + XS-HA?? Find starting moles of HA and moles of KOH. Starting moles HA: Starting moles KOH: Now, Let’s examine the resulting equilibrium. Therefore, we have 0.8 x 10 -3 moles of HA left and 2.2 x 10 -3 moles of its conjugate base (A - ) formed. We can now change these amounts to Molarity and use them in our equilibrium expression. HA + H 2 O ↔ H 3 O + + A - I. C. E. 0.0154 0 0.0423 -x +x +x 0.0154-x x 0.0423+x

9. A weak acid, HA (K a = 1.6 x 10 -5 ), is being titrated with a strong base, KOH. Determine the pH of the solution resulting from the mixing of 30.00 mL of 0.10 M HA with 12.00 mL of 0.10 M KOH. Now, Let’s examine the resulting equilibrium. Therefore, we have 1.8 x 10 -3 moles of HA left and 1.2 x 10 -3 moles of its conjugate base (A - ) formed. We can now change these amounts to Molarity and use them in our equilibrium expression. HA + H 2 O ↔ H 3 O + + A - I. C. E. 0.0154 0 0.0423 -x +x +x 0.0154-x x 0.0423+x Try dropping the -x and +x terms. If the value of x comes out to less than 5% of 0.0286, dropping the term is justified. x = [H 3 O + ] = 2.4 x 10 -5 pH = -log [5.825 x 10 -5 ] = 5.23