Acids & Bases

Solutions to Study Problems

Experiment IV:

Acids & Bases

1. Calculate the pH, [H+], and [OH-] of the following solutions of strong acids and basee:

a) 0.001M HCl d) 0.012 Ba(OH)2

Solution: 0.001M HCl = 0.001M H+ Solution: 0.024M = [OH-]

[OH-] = 1 x 10-11M 4.2 x 10-11M = [H+]

pH= -log (0.001M)=3.0 pH= -log(4.2 x 10-11)=12.0

pH= -log(1 x 10-11) = 11.0 pOH=-log(0.024)=1.6

b) 0.125M HNO3 e)2.1 x 10-4M HClO4

Solution: 0.125M= [H+] Solution: 2.1 x 10-4 M= [H+]

8 x 10-14= [OH-] 4.8 x 10-11 M= [OH-]

pH= -log(0.125)= 0.90 pH= 3.7

pOH= -log(8 x 10-14)= 13.1 pOH=10.0

c) 0.0031M NaOH

Solution: 0.0031M = [OH-]

3.2 x 10-12 M = [H+]

pH= -log(3.2 x 10-12)= 2.5

pOH= -log(0.0031)= 11.0

2. At body temperature (37û C) the ion product constant of water, Kw, is 2.42 x 10-14. What are the concentrations of H3O+ and OH- in pure water at 37 ûC?? What is the pH of pure water at this temperature??

Solution: Kw=[OH-][H+]=2.42 x 10-14

[H+]=[OH-]=1.56 x 10-7M

pH= -log(1.56 x 10-7) = 6.81

3) Identify the following as representing acidic, basic, or neutral solutions:

a) pH=3.54 Solution: acidic b) pH=8.25 Solution: basic

c) pOH=7.00 Solution: neutral d) pOH=10.43 Solution: acidic

e) pOH=2.25 Solution: basic

4)Calculate the [H+] and [OH-] in a solution having a pH equal to:

a) 1.30

Solution: 1.30= -log([H+])

[H+]= 0.050M

[OH-]= 2.00 x 10-13 M

b) 5.73

Solution: [H+] = 1.86 x 10-6M

[OH-]=5.38 x 10-9M

c) 4.00

Solution: [H+] = 1 x 10-4 M

[OH] = 1 x 10-10 M

d) 7.80

Solution: [H+] = 1.58 x 10-8 M

[OH-] = 6.33 x 10-7 M

e) 10.74

Solution: [H+] = 1.82 x 10-11 M

[OH-] = 5.49 x 10-4M

f) 12.61

Solution: [H+] = 2.45 x 10-13

[OH-] = 0.0408M

5) If 28.7mL of 0.375M NaOH is necessary to neutralize 40.0mL of an HCl solution, calculate the molarity of the HCl sol'n.

Solution: (0.0287L NaOH)(0.375 moles NaOH/L)=[HCl](0.040L HCl)

[HCl] = 0.253M

6) Sulfuric acid (H2SO4) can be neutralized with NaOH according to:

H2SO4(aq) + 2NaOH(aq) ------> Na2SO4(aq) + 2H2O

What volume of a 0.100M NaOH solution is necessary to neutralize 100.0mL of 0.050M H2SO4??

Solution: (0.100M NaOH)(volume of NaOH)=(0.100L H2SO4)(0.050M H2SO4)(2)

volume of NaOH=0.100L=100mL

7) Is it possible to have a pH other than 7.0 at the equivalence point in an acid-base titration?

Solution: When a weak acid is titrated with a strong base or a weak base is titrated with a strong acid, it is possible.

When a strong acid is titrated by a strong base or vice versa, we see a pH of 7.00.

8) In a titration, the equivalence point and the end point are ofen not exactly the same. Justify this statement.

Solution: At the equivalence point, by definition, the number of equivalents of acid or base added equals the number of equivalents of acid or base titrated. The end point is the moment in the titration at which your indicator believes that the equivalence point has been reached.

9) Ascorbic acid (Vitamin C) is a diprotic acid of molecular weight 176g/mole. A sample of a vitamin supplement was analyzed by titating a 0.100gram sample dissolved in water with 0.020M NaOH. A volume of 15.2mL of the base was required to completely neutalize the ascorbic acid. What was the weight percent ascorbic acid in the sample??

Solution: (0.0152L NaOH)(0.200moles NaOH/L) = 0.003moles NaOH needed to neutralize H+

(0.003 moles H+)/2=0.0015moles ascorbic acid (diprotic)

(0.0015 moles)(176 grams/mole)= 0.027 grams ascorbic acid

0.027 g ascorbic acid/0.1 grams sample = 27% ascorbic acid

10) Acetylsalicylic acid, commonly known as aspirin (C8H7O2COOH, MW=180g/mole), is a weak acid with a Ka of 3.4 x 10-4. A tablet containing 500mg of aspirin is dissolved in 100mL of water.

a) Calculate the pH of this solution.

Solution: 500mg of aspirin=0.5g aspirin(1 mole/180 g)=0.0028 moles aspirin

concentration of aspirin=0.0028 moles/0.1 L=0.028M

NOTE: aspirin is a weak acid so you cannot assume it dissociates completely

Ka=[H+][A-]/[HA]=3.4 x 10-4

initially, [HA]=0.028M, [H]=0M and [A]=0M

as the dissociation goes, [HA] changes by -x and [H] and [A] increase by x

finally, [HA]=0.028-x, [H]=x and [A]=x

so, the Ka=x2/0.028....so, x=[H+]=0.0031M

pH=2.50

b) How many mL of 0.1M NaOH are required to titrate the aspirin solution??

Solution: 0.1M NaOH(volume NaOH required)=(0.0028moles H+ from aspirin)

volume of NaOH needed=0.028L=28mL

c) What is the pH at the equivalence point??

Solution: After the salt has been neutralized, you have a lot of A-. The anion undergoes hydrolysis to reform HA.

Because you already know the Ka, and [HA] and [A-] are an acid base pair, you can say that the Kw=Ka + Kb

Kb=2.94 x 10-11=[OH][HA]/[A-]

=the constant associated with reassociation=

according to the reaction, A- + H2O -----> HA + OH-

2.94 x 10-11=x2/0.028

x=[OH]=3.31 M

pOH=6.04........pH=14.0-6.04=7.96