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Principles of Chemistry II © Vanden Bout Today

How to solve all acid/base problems

(except the ones we'll do next week)

Strong Acid

Weak Acid

Buffer

Weak Base

Strong Base

Principles of Chemistry II © Vanden Bout

If you're having trouble

go step by step

1. Remove the spectator ions (Na

, Cl , NO 3

,....)2. Are there any strong acids or bases3. Are there any weak acids or bases4. Do I neutralize?

(strong acid and any base. Strong base and any acid)

5. Neutralize

convert everything to moles write down the correct neutralization reaction find the limiting reagent final the compounds in the solution after neutralization convert back to molarity

6. Identify what is in solution and solve the equilibrium7. convert to the appropriate answer pH, pOH, [H

Principles of Chemistry II © Vanden Bout what is the pH of a solution of

100 mL of a 1M weak acid with a K

a of 10 -4 and 50 mL of 1M NaOH? spectator ion? Na (H ) strong acid? No (OH ) strong base? YES (HA or BH ) weak acid? YES (B or A ) weak base? No Do I need to neutralize? Yes. I have a strong base and an acid Principles of Chemistry II © Vanden Bout what is the pH of a solution of

100 mL of a 1M weak acid with a K

a of 10 -4 and 50 mL of 1M NaOH?

Convert to moles

.1L x 1M = 0.1 moles of HA .05L x 1M = 0.05 moles of OH neutralization reactionOH (aq) + HA(aq) A (aq) + H 2 O(l) init0.05 moles0.1 moles0 moleschange-0.05 moles-0.05 moles+0.05 molesafter neutral

0 moles0.05 moles0.05 molesICAN

not ICE Principles of Chemistry II © Vanden Bout what is the pH of a solution of

100 mL of a 1M weak acid with a K

a of 10 -4 and 50 mL of 1M NaOH? neutralization reactionOH (aq) + HA(aq) A (aq) + H 2

O(l) after

neutral

0 moles0.05 moles0.05 molesWhat "kind" of solution is this?buffer[HA] =0.05/.15 = .333[A

] =0.05/.15 = .333 Principles of Chemistry II © Vanden Bout How do I calculate the pH for different solutions?Strong Acid [H ] = C a

Strong Base

[OH ] = C b

Weak Acid

[H ] = sqrt(K a C a

Weak Base

[OH ] = sqrt(K b C b

Buffer

[H ] =K a C a /C b

Buffer

[OH ] =K b C b /C a Principles of Chemistry II © Vanden Bout what is the pH of a solution of

100 mL of a 1M weak acid with a K

a of 10 -4 and 50 mL of 1M NaOH? [HA] =0.05/.15 = .333[A ] =0.05/.15 = .333Buffer [H ] =K a C a /C b

Buffer

[H ] =10 -4 (.333)/(.333) = 10 -4 pH = 4pOH = 10

Principles of Chemistry II © Vanden BoutWhat is the pH of a solution formed by mixing

100 mL of Acetic Acid and 100 mL of LiOH?

A. a strong acid and a strong base B. a weak acid and a strong base C. a weak acid and a weak base D. a strong acid and a weak base E. a strong acid and a weak acid

Acetic Acid = HA

NaOH = OH

what acid/base species are in this solution to start?

Principles of Chemistry II © Vanden BoutWhat is the pH of a solution formed by mixing

100 mL of Acetic Acid and 100 mL of LiOH?

A. yes B. no C. it depends on the molecular weight of acetic acid Do I need to neutralize this solution?Strong base and an acid OH + HA A H 2 O

Principles of Chemistry II © Vanden BoutWhat is the pH of a solution formed by mixing

100 mL of Acetic Acid and 100 mL of LiOH?

A. HA B. A C. of HA and of A

D. of OH

What is left in solution after neutralization?.1 L x 1M = .1 HA .1L x 1M = .1 OH note: this would be the equivalence point in a titration

Principles of Chemistry II © Vanden BoutWhat is the pH of a solution formed by mixing

100 mL of Acetic Acid and 100 mL of LiOH?

A. strong acid B. weak acid C. buffer

D. weak base

E. strong base

What "kind" of equilibrium problem is this?

A is a weak base no HA in solution no H no OH

Principles of Chemistry II © Vanden BoutWhat is the pH of a solution formed by mixing

100 mL of Acetic Acid and 100 mL of LiOH?

A. 10 -5 B. 10 -7 C. 10 -9

D. 10

-14

If the K

a of acetic acid 10 -5 what is K b for acetate (A K b for A = K w /K a for HA 10 -14 /10 -5 = 10 -9

Principles of Chemistry II © Vanden BoutWhat is the pH of a solution formed by mixing

100 mL of Acetic Acid and 100 mL of LiOH?

A. 3.8 B. 4.65 C. 7

D. 9.35

E. 11.2

What is the pH of this solution given that we end up with

0.1 moles of A

with a K b of 10 -9

It is a base so it should be > 7

C b = .1/.2 = 0.5 M [OH ] = sqrt(K b C b ) = sqrt(5x10 -1 x10 -9 sqrt(5x10 -10 )=2.23x10 -5 pOH =4.65 pH = 9.35 Principles of Chemistry II © Vanden Bout

How to get a bufferStart with HA and A

or BH and B partially neutralize HA with OH this will generate A partially neutralize B with H this will generate BH Principles of Chemistry II © Vanden Bout

Which of the following can make a buffer?

A. a strong acid and a weak acid B. a strong acid and a strong base C. a strong acid and a weak base

D. a strong base and a weak base

H neutralizes B to make some BH Principles of Chemistry II © Vanden Bout mixing equal volumes of equal concentrations solutions which of the following is a buffer? A. HNO 3 and NaNO 3 B.

HCl and KCl

C.

HF and NaF

D. HClO

4 and CsClO 4

HA = HF

A = F

HA and A

in the same solution buffer Principles of Chemistry II © Vanden Bout

Let's look at some examples

Principles of Chemistry II © Vanden Bout Rolaids® contain about of Magnesium Hydroxide Why in the world would you ever put such a thing in your mouth? A. is nothing. I each 10 NaOH daily just for laughs B.

Acids are dangerous by bases as quite safe

C. The saliva in my mouth is acidic enough to "handle it" D.

Mg(OH)

2 is not soluble in water Principles of Chemistry II © Vanden Bout

Solubility EquilibriaMg(OH)

2 (s) Mg 2+ (aq) + 2OH (aq) K sp = [Mg 2+ ][OH 2 = 5.6 x 10 -12OH that is dissolved neutralizes any H then more OH dissolves...repeat end result is a very slightly basic solutionquotesdbs_dbs20.pdfusesText_26