Mixing Acids and Bases Practice II -- Chemistry 121A Hanson

[Pages:2]Mixing Acids and Bases Practice II -- Chemistry 121A Hanson

When we mix an acid and a base we expect a reaction. As such, our task in predicting the final pH of the solution is to:

a) Identify the principal acidic and basic species being mixed.

b) Write a net ionic equation leading to products (conjugate acid and base).

c) Determine the number of moles of whatever is present after the "stoichiometric reaction."

d) If necessary, write a second net ionic equation for the reaction of one or more of those species with water.

e) Determine the initial concentrations needed for an equilibrium calculation. f) Carry out a calculation (possibly using an ICE table) to determine [H3O+] or [OH-] at equilibrium.

g) From that, calculate the pH.

Yes, that's rather involved. But you can do it! You have all the tools you need. What you can see, I am sure, is that all the concepts we have been working on ? weak and strong acids and bases, principal species in solution, net ionic equations, molarity, limiting reactants, equilibrium algebra, definitions of pH and pOH ? these are all critical components of success in this area. You have learned or are very close to have learned all these things. Now we get to apply them, both in class and in lab. This is what we will be doing for the rest of the semester.

Suggested on-line reading includes:

Principal Species in Solution ? a discussion of principal species

Acid-Base Mixing Calculator ? dial in whatever combination of acids and bases you want and see how the pH changes as they are mixed.

In each case, determine the pH of the solution. The first two are just review.

1. A 0.20 M solution of HCl.

2. A solution containing 0.20 moles of HCl in 750 mL of water.

3. A solution prepared by adding 0.20 moles of HCl and 0.20 moles of NaF to enough water to make 750 mL.

4. A solution prepared by mixing 50 mL of 0.20 M NaF and 100 mL of 0.20 M HCl.

5. A solution prepared by mixing 50 mL of 0.20 M NaF and 100 mL of 0.10 M HCl.

6. A solution prepared by mixing 100 mL of 0.20 M NaOH and 50 mL of 0.10 M HF.

7. A solution prepared by mixing 50 mL of 0.20 M NaOH and 100 mL of 0.10 M HF.

Think about how you might do this one, although really it is the subject of the next worksheet:

8. A solution prepared by mixing 25 mL of 0.20 M NaOH and 100 mL of 0.10 M HF.

Solutions:

1. (a) H3O+ is the principal species and pH = -log (0.20) = 0.70

2. (a) H3O+ is the principal species and pH = -log (0.20/0.750) = 0.57

3. (a) H3O+ and F- are the principal species.

(b) stoichiometric reaction is H3O+ + F- H2O + HF

(c) 0.20 moles of H3O+ and 0.20 moles of F- react completely to form 0.20 mol of HF

(d) equilibrium with water is HF + H2O

F- + H3O+ with K = Ka = 3.5 x 10-4

(e) [HF]o = 0.20 mol / 0.750 L = 0.27 M.

(f) algebra gives x2/(0.27 ? x) = 3.5 x 10-4 where [H3O+] = x = 0.0095476, and pH = 2.02

4. (a) H3O+ and F- are the principal species. (b) stoichiometric reaction is H3O+ + F- H2O + HF (c) 20 mmol of H3O+ and 10 mmol of F- react to give 10 mmol of HF with 10 mmol of H3O+ remaining (d) Ignore the weak acid HF because of the presence of H3O+ (e) [H3O+] = 10 mmol / 150 mL = 6.67 x 10-2 M and pH = 1.18

5. (a) H3O+ and F- are the principal species.

(b) stoichiometric reaction is H3O+ + F- H2O + HF

(c) 10 mmol of H3O+ and 10 mmol of F- react completely to give 10 mmol of HF

(d) equilibrium with water is HF + H2O

F- + H3O+ with K = Ka = 3.5 x 10-4

(e) [HF]o = 10 mmol / 150 mL = 6.67 x 10-2 M

(f) algebra gives x2/(6.67 x 10-2 ? x) = 3.5 x 10-4 where [H3O+] = x = 4.65863E-3, and pH = 2.33

6. (a) OH- and HF are the principal species. (b) stoichiometric reaction is HF + OH- F- + H2O (c) 5 mmol of HF reacts with 20 mmol of OH- to give 5 mmol of F- with 15 mmol of OH- remaining (d) ignore weak base in the presence of OH- (e) [OH-] = 15 mmol / 150 mL = 0.10 M, pOH = 1.00, and pH = 13.00

7. (a) OH- and HF are the principal species.

(b) stoichiometric reaction is HF + OH- F- + H2O

(c) 10 mmol of HF reacts with 10 mmol of OH- completely to give 10 mmol of F-

(d) equilibrium with water is F- + H2O

HF + OH- with K = Kw/Ka(HF) = 2.86 x 10-11

(e) [F-]o = 10 mmol / 150 mL = 6.67 x 10-2 M

(f) algebra gives x2/(6.67 x 10-2 ? x) = 2.86 x 10-11 where [OH-] = x = 1.38046E-6, and pH = 8.14

8. (a) OH- and HF are the principal species.

(b) stoichiometric reaction is HF + OH- F- + H2O

(c) 10 mmol of HF reacts with 5 mmol of OH- to give 5 mmol of F- with 5 mmol of HF remaining

(d) equilibrium with water is HF + H2O

F- + H3O+ with K = Ka = 3.5 x 10-4

(e) [HF]o = [F-]o = 5 mmol / 125 mL = 0.040 M; we approximate [H3O+] = 0 to start.

(f) algebra gives (0.040 + x)x/(0.040 ? x) = 3.5 x 10-4 where [H3O+] = x = 3.44031E-4, and pH = 3.46

Note that #8 is a buffer and that we could have used the Henderson-Hasselbalch equation:

pH = pKa + log([F-]/[HF])

with the approximation that [F-]/[HF] [F-]o/[HF]o = (mmol F-)/(mmol HF) = 1.0 in this case. Thus in this case pH = pKa + log(1.0) = -log(3.5 x 10-4) + 0 = 3.46, as above. So we could have solved this a bit more quickly:

8. (a) OH- and HF are the principal species. (b) stoichiometric reaction is HF + OH- F- + H2O (c) 10 mmol of HF reacts with 5 mmol of OH- to give 5 mmol of F- with 5 mmol of HF remaining (d) pH = pKa + log((mmol F-)/(mmol HF)) = -log(3.5 x 10-4) + log(1.0) = 3.46

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