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Updates. Midterms marked; solutions are posted Assignment 03 is in the box Assignment 04 is up on ACME and is due Mon., Feb. 26 (in class). Acids and Bases. Chapter 16. How do we measure pH?. For less accurate measurements, one can use Litmus paper Turns blue above ~pH = 8

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  1. Updates • Midterms marked; solutions are posted • Assignment 03 is in the box • Assignment 04 is up on ACME and is due Mon., Feb. 26 (in class)

  2. Acids and Bases Chapter 16

  3. How do we measure pH? • For less accurate measurements, one can use • Litmus paper • Turns blue above ~pH = 8 • Turns red below ~pH = 5 • An indicator

  4. How do we measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

  5. Strong acids • You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. • These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. • For the monoprotic strong acids, [H3O+] = [acid].

  6. Strong bases • Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+). • Again, these substances dissociate completely in aqueous solution.

  7. Kc = [H3O+] [A−] [HA] HA(aq) + H2O(l) A−(aq) + H3O+(aq) Dissociation constants • For a generalized acid dissociation, the equilibrium expression would be • This equilibrium constant is called the acid-dissociation constant, Ka.

  8. Dissociation constants The greater the value of Ka, the stronger the acid.

  9. HF (aq) H+(aq) + F-(aq) = 7.1 x 10-4 = 7.1 x 10-4 = 7.1 x 10-4 [H+][F-] x2 x2 Ka Ka = Ka = 0.50 - x 0.50 [HF] HF (aq) H+(aq) + F-(aq) What is the pH of a 0.5M HFsolution (at 250C)? Initial (M) 0.50 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.50 - x x x Ka << 1 0.50 – x 0.50 x2 = 3.55 x 10-4 x = 0.019 M pH = -log [H+] = 1.72 [H+] = [F-] = 0.019 M [HF] = 0.50 – x = 0.48 M 16.5

  10. = 7.1 x 10-4 0.006 M 0.019 M x2 x 100% = 12% x 100% = 3.8% 0.05 M 0.50 M Ka 0.05 When can I use the approximation? Ka << 1 0.50 – x 0.50 When x is less than 5% of the value from which it is subtracted. Less than 5% Approximation ok. x = 0.019 What is the pH of a 0.05M HFsolution (at 250C)? x = 0.006 M More than 5% Approximation not ok. Must solve for x exactly using quadratic equation. 16.5

  11. Solving weak acid ionization problems: • Identify the major species that can affect the pH. • In most cases, you can ignore the autoionization of water. • Ignore [OH-] because it is determined by [H+]. • Use ICE to express the equilibrium concentrations in terms of single unknown x. • Write Kain terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. • Calculate concentrations of all species and/or pH of the solution. 16.5

  12. = 5.7 x 10-4 = 5.7 x 10-4 0.0083 M x2 x2 x 100% = 6.8% 0.122 M Ka Ka = 0.122 - x 0.122 HA (aq) H+(aq) + A-(aq) What is the pH of a 0.122M monoprotic acid whose Ka is 5.7 x 10-4? Initial (M) 0.122 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.122 - x x x Ka << 1 0.122 – x 0.122 x2 = 6.95 x 10-5 x = 0.0083 M More than 5% Approximation not ok. 16.5

  13. -b ± b2 – 4ac = 5.7 x 10-4 x = 2a x2 Ka = 0.122 - x HA (aq) H+(aq) + A-(aq) Initial (M) 0.122 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.122 - x x x x2 + 0.00057x – 6.95 x 10-5 = 0 ax2 + bx + c =0 x = 0.0081 x = - 0.0081 pH = -log[H+] = 2.09 [H+] = x = 0.0081 M 16.5

  14. Ionized acid concentration at equilibrium x 100% x 100% Percent ionization = Initial concentration of acid [H+] [HA]0 percent ionization = For a monoprotic acid HA [HA]0 = initial concentration 16.5

  15. [H3O+] [COO−] [HCOOH] Ka = Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • We know that

  16. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • To calculate Ka, we need the equilibrium concentrations of all three things. • We can find [H3O+], which is the same as [HCOO−], from the pH.

  17. Calculating Ka from the pH pH = −log [H3O+] 2.38 = −log [H3O+] −2.38 = log [H3O+] 10−2.38 = [H3O+] 4.2  10−3 = [H3O+] = [HCOO−]

  18. Calculating Ka from pH Now we can set up a table…

  19. [4.2  10−3] [4.2  10−3] [0.10] Ka = Calculating Ka from pH = 1.8  10−4

  20. Polyprotic Acids • Have more than one acidic proton. • If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

  21. NH3(aq) + H2O (l) NH4+(aq) + OH-(aq) [NH4+][OH-] Kb = [NH3] weak base strength Solve weak base problems like weak acids except solve for [OH-] instead of [H+]. Kb Weak Bases and Base Ionization Constants Kb is the base ionization constant 16.6

  22. 16.6

  23. HA (aq) H+(aq) + A-(aq) A-(aq) + H2O (l) OH-(aq) + HA (aq) H2O (l) H+(aq) + OH-(aq) Kw Kw Ka= Kb= Kb Ka Ionization Constants of Conjugate Acid-Base Pairs Ka Kb Kw KaKb = Kw Weak Acid and Its Conjugate Base 16.7

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