1 / 24

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 + ]

hieu
Download Presentation

pH and [H+]

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related