Acid Base Stoichiometry and Titrations

1. Acid Base Stoichiometry and Titrations By Marisa Carlson and Lizzie Sokol

2. Stoichiometry • In an acid, HxA, or base, B(OH)x, x determines the number of equivalence points in the titration.

3. What is a Titration? • A procedure for quantitative analysis of a substance by an essentially complete reaction in solution with a measured quantity of a reagent of known concentration.

4. Finding Equivalence Points • n1V1C1 = n2V2C2 • n is the number of moles of H+ or OH- in the acid and base to be titrated and/or to titrate • C is the concentration of the acid and base • V is the volume, one quantity is given, the other one is unknown and determines how much titrant is needed to reach the first equivalence point (all number relating to acid go on one side, and all relating to base go on the other)

5. A Sample Titration Problem • 20.00 mL of 0.250 M H2SeO3 is titrated with 0.250 M NaOH • Ka1 = 2.7 x 10-3 • Ka2 = 2.5 x 10-7 • Find pH when 0, 10, 17, 20, and 45 mL of NaOH have been added

6. Zero mL H2SeO3 + H20 ↔ HSeO3- + H3O+ I .250M*20mL 0 0 5mmol C -x +x +x E 5mmol/20mL x x ≈.250M

7. Zero mL con’t Ka1 = [HSeO3-]*[H3O+]/[H2SeO3] 2.7 x 10-3 = x*x/.250 x=.026 M -log(x) = pH = 1.59

8. Ten mL • At the half-way point of the titration the pKa = pH • So… -log (2.7x10-3) = 2.57

9. Seventeen mL H2SeO3 + H20 ↔ HSeO3- + H3O+ I 5mmol 0 0 -(17mL*.250M) + (17mL*.250M) .750mmol 4.25mmol C -x +x +x E ≈.750mmol ≈4.25mmol x

10. Seventeen mL con’t Ka1 = [HSeO3-]*[H3O+]/[H2SeO3] 2.7 x 10-3 = 4.25*x/.750 x=4.76x10-4 M -log(x) = pH = 3.32

11. Twenty mL • At each equivalence point before the end is reached, it must be determined whether the Ka or the Kb should be used. • Ka2 = 2.5x10-7 • Kb1 = 1x10-14 / 2.7x10-3 = 3.7x10-12 • So, the acidic properties are stronger, and the solution continues as an acid (Ka>Kb)

12. Twenty mL con’t H2SeO3 + H20 ↔ HSeO3- + H3O+ I 5mmol 0 0 -5mmol +5mmol 0 5mmol Because all of the H2SeO3 is used up, and it acts as an acid, we move on the the next equation… HSeO3- + H20 ↔ SeO32- + H3O+

13. Twenty mL con’t HSeO3- + H20 ↔ SeO32- + H3O+ I 5mmol 0 0 C -x +x +x E 5mmol/40mL .125M x x

14. Twenty mL con’t Ka2 = [SeO32-]*[H3O+]/[HSeO3-] 2.5 x 10-7 = x*x/.125 x=1.77x10-4 M -log(x) = pH = 3.75

15. Forty-five mL • Since all of the acid has been neutralized (beyond last equivalence point), the calculation does not involve neutralization. 5mL NaOH (not used to neutralize acid) *.250M = 1.25mmol 1.25 mmol / 65 mL = .0192 M -log (.0192) = pOH = 1.72 14 – pOH = pH = 14 – 1.72 = 12.28

16. Helpful Hints • You can use the concentration of the acid or base in the ice chart as long as the concentrations of both products are unknown. • pH= pKa + log([A-]/[HA]) • pOH= pKb + log ([HB+]/[B]) • Remember that the pH of a solution after the acid or base has been completely neutralized does not require an ice chart. • Determining whether the Ka or Kb should be used requires finding both and finding the larger quantity.

17. Titrating a Base with an Acid • Virtually no difference except: • when solving for x, x= [OH-] • so, when the -log(x) is taken, the result is the pOH • therefore, to find the pH, subtract pOH from 14