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Chemistry:

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.

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Chemistry:

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  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 + H2O ↔ H3O+ + 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 = [H3O+] = 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 Kb of the B is 1.4 x 10-5 . Chemistry: B + H2O ↔ 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. Ka1 of the H2C6H6O6 is 6.8 x 10-5, Ka2 = 2.8 x10-12 . Chemistry: H2C6H6O6 + H2O ↔ H3O+ + HC6H6O6- 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 = [H3O+] = 1.17 x 10-4 pH = -log[x] = 3.94

  4. A weak acid, HA (Ka = 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) → H2O(l) + KA(aq) + XS-HA?? Find starting moles of HA and moles of KOH. Starting moles HA: Starting moles KOH: 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. Now, Let’s examine the resulting equilibrium. We can now change these amounts to Molarity and use them in our equilibrium expression.

  5. A weak acid, HA (Ka = 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) → H2O(l) + KA(aq) + XS-HA?? Find starting moles of HA and moles of KOH. Starting moles HA: Starting moles KOH: 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. Now, Let’s examine the resulting equilibrium. We can now change these amounts to Molarity and use them in our equilibrium expression. HA + H2O ↔ H3O+ + A- 0.0429 0 0.0286 I. C. E. -x +x +x 0.0429-x x 0.0286+x

  6. A weak acid, HA (Ka = 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. 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. Now, Let’s examine the resulting equilibrium. We can now change these amounts to Molarity and use them in our equilibrium expression. HA + H2O ↔ H3O+ + A- 0.0429 0 0.0286 I. C. E. -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 = [H3O+] = 2.4 x 10-5 pH = -log [2.4 x 10-5] = 4.62

  7. A weak acid, HA (Ka = 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) → H2O(l) + KA(aq) + XS-HA?? Find starting moles of HA and moles of KOH. Starting moles HA: Starting moles KOH: 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. Now, Let’s examine the resulting equilibrium. We can now change these amounts to Molarity and use them in our equilibrium expression.

  8. A weak acid, HA (Ka = 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) → H2O(l) + KA(aq) + XS-HA?? Find starting moles of HA and moles of KOH. Starting moles HA: Starting moles KOH: 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. Now, Let’s examine the resulting equilibrium. We can now change these amounts to Molarity and use them in our equilibrium expression. HA + H2O ↔ H3O+ + A- 0.0154 0 0.0423 I. C. E. -x +x +x 0.0154-x x 0.0423+x

  9. A weak acid, HA (Ka = 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. 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. Now, Let’s examine the resulting equilibrium. We can now change these amounts to Molarity and use them in our equilibrium expression. HA + H2O ↔ H3O+ + A- 0.0154 0 0.0423 I. C. E. -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 = [H3O+] = 2.4 x 10-5 pH = -log [5.825 x 10-5] = 5.23

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