1. Acid-Base Problems

2. A complete forward reaction or equilibrium? If the problem involves only strong acids and/or bases, the reaction goes to completion (→). If the problem involves a weak acid or base, the reaction involves equilibrium (↔).

3. If the problem involves only strong acids and/or bases, the reaction goes to completion (→). If only strong acid or base, use one of these four relationships to solve: – pH = -log [H + ]pOH = -log[OH - ] – [H + ][OH - ] = K w pH + pOH = 14 For titration (acid + base), use regular stoichiometry to find molarity, then use log formula to find pH if needed. (or use pH to find molarity depending on given.)

4. If the problem involves a weak acid or base, the reaction involves equilibrium (↔). Use RICE table. – Weak acids hydrolyze to produce H + and a conjugate base. K a is used. – Weak bases hydrolyze to produce OH - and a conjugate acid. K b is used. For titration (acid + base), two steps are needed: 1.Use regular stoichiometry to find which species are left after neutralization. 2.Use RICE table and ionization constant (K a or K b ) to solve with remaining species. – Note that K a is needed if H + is a product in the equilibirium and K b is needed if OH - is a product.

5. To do all titration problems: I. Start with stoichiometry: Write balanced equation for reaction. Find # moles H + (or OH - ) you are starting with. If amount of titrant is given, find how many moles of that you used. Use stoichiometry to figure out what is left over. (If you have more H + than OH -, then all OH - will be converted to water.) Subtract the # moles used from the dominant species. The # moles water will also result in the “liberation” of the same # moles of the conjugate.

6. II. Analyze what is left after neutralization. If only base is left, use K b and write the ICE table as basic. If only acid is left, use K a and write ICE table as hydrolysis of the H +. III. Do equilibrium problem. (ICE table) Do not forget to include the starting concentration of the conjugate (A - if you started with HA).

7. Special case ½ way to equivalence point, pH = pK a (This is a short cut!) At equivalence point, all of the H + is neutralized. Note: HA + H 2 O   H 3 O + + A - I will write it in this shortened form: HA   H + + A -

8. A 27.1 mL sample of a 0.412 M aqueous hydrocyanic acid solution is titrated with a 0.444 M aqueous barium hydroxide solution. What is the pH at the start of the titration, before any barium hydroxide has been added? Nothing has been added, so it is not a titration yet. Solve like any equilibrium problem: HCN  H + + CN -.412 M 00 K a = x 2 = 6.2 x 10 -10 -x+x+x.412.412-xxxx = [H + ] = 1.598 x 10 -5 M pH = -log[H + ] = 4.80

9. A 41.9 mL sample of a 0.475 M aqueous hydrocyanic acid solution is titrated with a 0.422 M aqueous barium hydroxide solution. What is the pH after 10.0 mL of base have been added? Stoichiometry: 2HCN + Ba(OH) 2  2H 2 O + Ba(CN) 2 Ba(CN) 2 is water soluble. (.0419 L)(.475 mol/L HCN) =.0199 mol HCN (so.0199 mol H + is available) (.010L)(.422 mol Ba(OH) 2 )(2mol OH - ) =.00844 mol OH - LL

10. More H + than OH - :. 0199 mol H + -.00844 mol OH - =.01146 mol HCN left (&.00844 mol CN - liberated) Analyze what is left: Total volume =.0419 L +.010 L =.0519 LH 2 O, HCN and CN - are present Concentrations: H 2 O = N/A [HCN] =.01146mol =.221 M [CN - ] =.00844 mol =.163 M.0519 L.0519 L Since acid conc is higher, use K a. Equilibrium: HCN  H + + CN -.221 M0.163K a = x(.163) = 6.2 x 10 -10 -x+x +x.221.221-xx.163 + xx = [H + ] = 8.4 x 10 -10 M pH = -log[H + ] = 9.08

11. When a 29.5 mL sample of a 0.497 M aqueous hydrofluoric acid solution is titrated with a 0.334 M aqueous potassium hydroxide solution, what is the pH at the midpoint in the titration? At midpoint, pH = pK a, so pH = -log (7.2 x 10 -4 ) = 3.14 You could prove this by finding that it would take.0439 mol OH - added to reach equivalence, divide by 2 to get 21.945 mL added at midpoint. Then do ICE to get [H + ] = 7.2 x 10 -4 M.

12. What is the pH at the equivalence point in the titration of a 19.0 mL sample of a 0.353 M aqueous hydrocyanic acid solution with a 0.408 M aqueous potassium hydroxide solution? Stoichiometry: 2HCN + Ba(OH) 2  2H 2 O + Ba(CN) 2 Ba(CN) 2 is water soluble. (.010 L)(.353 mol/L HCN) =.00671 mol H + all neutralized by KOH, so:.00671 mol OH - | 1 L =.0164 L KOH used at equivalence |.408 mol KOH Total vol =.0164 +.019 =.0354L

13. Equivalence point:.00671 mol H + +.00671 mol OH - = 0 mol HCN left (&.00671 mol CN - liberated) Analyze what is left: Total volume = =.0164L +.019L =.0354L H 2 Oand CN - are present Concentrations: H 2 O = N/A [CN - ] =.00671 mol =.190 M CN -.0354 L Since only a conjugate base is present, use K b : K b = 1 x 10 -14 = 1.61 x 10 -5 6.2 x 10 -10 Equilibrium: CN - + H 2 O  HCN + OH -.190 M 0 0K b = x 2 = 1.61 x 10 -5 -x +x +x.190.190-x x xx = [OH - ] = 8.5 x 10 -5 M pOH = -log[OH - ] = 4.07 pH = 14 – pOH = 9.93

14. When a 19.4 mL sample of a 0.382 M aqueous hydrocyanic acid solution is titrated with a 0.413 M aqueous barium hydroxide solution, what is the pH after 13.5 mL of barium hydroxide have been added? Stoichiometry: 2HCN + Ba(OH) 2  2H 2 O + Ba(CN) 2 Ba(CN) 2 is water soluble..0194L (.382 mol/L HCN) =.0074 mol H +.0135L|.413 mol Ba(OH) 2 | 2 mol OH - =.01115 mol OH - | L |1 mol Ba(OH) 2 There are more OH - ions, so the solution is basic: 01115 moles OH - -.0074 mol H + =.00375 mol OH - left and.0074 mol CN - hanging around.

15. More OH - than H + :.00375 mol OH - left and.0074 mol CN - liberated) Analyze what is left: Total volume = =.0194L +.0135L =.0329 L OH -, H 2 O and CN - are present [OH - ] =.00375 mol/.0329 L =.114 M OH - Concentrations: H 2 O = N/A [OH - ] =.00375 mol/.0329 L =.114 M OH - [CN - ] =.0074 mol =.225 M CN -.0329 L BUT…CN - is a weak base, and the hydroxide ions it will generate in water will be very few relative to the amount from the strong base used in titration. Relative to the OH -, the CN - is negligible, so pOH = -log [OH - ] =.94 pH = 14 -.94 = 13.06