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pH and [H+]

pH and [H+]. Using concentrations to determine pH, pOH , pKw and other interesting things!. pH Scale. Because of the tremendous range of [H + ] and [OH - ], scientists rely on the pH scale to communicate concentrations Equation for pH: pH = -log [H + ]

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pH and [H+]

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  1. pH and [H+] Using concentrations to determine pH, pOH , pKw and other interesting things!

  2. pH Scale • Because of the tremendous range of [H+] and [OH-], scientists rely on the pH scale to communicate concentrations • Equation for pH: pH = -log [H+] • We can also solve for [H+] given the pH: [H+] = 10-pH

  3. pOH and pKw • As seen before, the p in pH stands for “-log” • We can use this to find the pOH: pOH = -log [OH-] • Or we can use it to find pKw: pKw = -log [Kw] Note: Kw = 1.0 x 10-14 *ALWAYS!!!!

  4. pOH and pKw (continued) • Since Kw = [H+] [OH-], and • pH + pOH = pKw, and • pH + pOH = 14 • Example: What is the pOH of a pH = 6.4 solution? pH + pOH = 14 6.4 + pOH = 14 Therefore, pOH = 7.6

  5. Measuring pH • We can measure pH by electronic means using a pH meter; or • We can use some plant compounds and synthetic dyes that change colour when mixed with acid or base • These are called ACID-BASE INDICATORS • Ex. Litmus paper – red turns blue at 8.3, and red at 4.7 • See Page 609 Table 7: Acid-base indicators • What do you notice? Which would be used?

  6. pH of strong acids and bases • Strong acids: • HClO4 • HI • HBr • HCl • HNO3 • H3O+ See examples of calc’s p. 545 • Strong Bases • NaOH • LiOH • KOH • OH- See examples of calc’s p. 547

  7. 8.2 Weak acids and bases • Weak acids is an electrolyte that does not ionize completely in water to form H+ • Most common acids are weak • Ex. Vinegar – CH3COOH • Weak bases are electrolytes that do not ionize completely, and have a weak attraction for H+ • Ex. Ammonia - NH3

  8. General formula for Weak Acid HA + HOH  A- + H3O+ Acid + water  C. Base + C. Acid

  9. General formula for weak base B- + HOH  OH- + HB+ Base + water  c. Base + c. acid

  10. Percent ionization • Weak acids ionize little in aqueous solutions. • It usually ionizes less than 50%. • Their pH is close to 7 (neutral). • We can calculate this by using: • [H+] = p/100 x [HA] • Where p = percent ionization • [HA] is the concentration of the acid

  11. Percent ionization (continued) • Calculating Percent Ionization example: • The pH of a 0.10M methanoic acid solution is 2.38. Calculate percent ionization. • 1. Write out equation: HCO2H  H+ + HCO2- • 2. Write out what is given: [HCO2H] = 0.10 M; pH = 2.38 • 3. Calculate [H+] using pH: • [H+] = 4.2 x 10-3 M Step 4. Use % ionization equation: [H+] = p/100 x [HA] p = 4.2%

  12. Ionization constants for weak acids, Ka • Acid ionization constant, Ka, is the equilibrium constant for the ionization of an acid • For any acid, Ka is simply: • Ka = [products]a [reactants]b Where a and b are coefficients from the balanced ionization for the weak acid

  13. Ionization constants for weak bases, Kb • Base ionization constant, Kb, is the equilibrium constant for the ionization of a weak base • For any base, Kb is simply: • Kb = [products]a [reactants]b Where a and b are coefficients from the balanced ionization for the weak base

  14. Relating Ka and Kb • Kw = Ka x Kb • Since Kw = [H+] [OH-] • And Ka x Kb = [H+] [OH-] Calculating Ka from Kb: The Kb for N2H4, rocket fuel, is 1.7 x 10-6. What is the Ka? Kw = 1.0 x 10-14, Kb = 1.7 x 10-6; so KaKb = Kw Ka(1.7 x 10-6) = 1.0 x 10-14; Ka = 5.9 x 10-9

  15. Relating Ka and Kb (continued) • What is the value of Kb for C2H3O2-? • We must be given a Ka...but where!??!?!? • OH! Appendix. C9, p. 803 Table of Ka Values • Therefore, Ka for acetic acid = 1.8 x 10-5; and Kw = 1.0 x 10-14; now solve! KaKb = Kw (1.8 x 10-5) Kb = 1.0 x 10-14 Kb = 5.6 x 10-10 Note: There are no units for any of the K values!!

  16. Conjugate acids and bases • In general, there are a few rules: • The c. Base of a strong acid is a very w. Base. • The c. Base of a w. Acid is a w. Base. • The c. Base of a very weak acid is a s. Base. • The c. Acid of a very weak base is a strong acid. • The c. Acid of a weak base is a weak acid. • The c. Acid of a strong base is a very weak acid.

  17. ph of WEAK ACID SOLUTIONS(DOING ICE CHARTS!!) • Calculating [H+] and pH of a weak acid, given Ka. • Calculate [H+] and pH of a 0.10M acetic acid solution. Ka = 1.8 x 10-5 1. Write the balanced equation HC2H3O2 H+ + C2H3O2-; Ka = 1.8 x 10-5 2. Write the equilibrium expression, Ka: Ka = [H+] [C2H3O2-] [HC2O3O2] 3. Complete an ICE Chart

  18. ph of weak acids (continued) HC2H3O2 H+ + C2H3O2- I 0.10 0 0 C -x +x +x E 0.10-x x x Before we begin, check for the HUNDRED Rule: [HA]initial = 0.10 = 5.6 x 103>>100; so, Ka 1.8 x 10-5 Therefore, assume x << 0.10; remove (-x)

  19. ph of weak acids (continued) • Plug into expression: • Ka = [H+] [C2H3O2-] = 1.8 x 10-5 [HC2H3O2] So: 1.8 x 10-5 = x2x = 1.3 x 10-3 0.10 Therefore: [H+] = 1.3 x 10-3M; and pH = 2.89

  20. Solving for Ka given pH • The pH of a 0.100M HOCl is 4.23. What is the Ka? So: pH = 4.23; so [H+] = 10-4.23 = 5.9 x 10-5M • Write the equation: HOCl H+ + OCl- • Write the expression: Ka = [H+] [OCl-] [HOCl] 3. Note: 1 mole of HOCl = 1 mole of H+; so, [H+] = [HOCl] = 5.9 x 10-5: so plug in now! 4. Ka = (5.9 x 10-5)2 = 3.5 x 10-8 0.100

  21. ph of weak base solutions • To calculate Kb you will have to: • Determine the formula of the base’s c. Acid (add an H+ to the base) • Locate the formula of the c. Acid in the chart • Note the c. Acid’s Ka • Substitute the Ka value into KaKb = Kw and solve for Kb • Use that Kb value to solve just as you did for Ka

  22. ph of weak base solutions (continued) • Finding the Kb, pH and pOH for weak bases Example: Calculate the pH of a 0.100 M N2H4 solution. The Kb = 1.7 x 10-6. 1. Write the equation(include H2O with w. base) N2H4 + HOH  N2H5+ + OH-; Kb = 1.7 x 10-6 2. Write the equilibrium expression: Kb = [OH-] [N2H5+] [N2H4] 3. Do ICE Chart!

  23. ph of weak base solutions (continued) • N2H4 + HOH  N2H5+ + OH- I 0.100 - 0 0 C -x - +x +x E 0.100-x - x x Check for hundred rule – yes it applies!!! 4. Plug in and solve: Kb = x2 = 1.7 x 10-6 x = 4.12 x 10-4 0.100

  24. ph of weak base solutions (continued) • BUT: We were asked for pH and need H+, not OH-... • So: We can change [OH-]  pOH • pOH = -log [OH-]; pOH = 3.38 • Remember: pH + pOH = 14; so • pH + 3.38 = 14; pH = 10.62

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