CHEMISTRY 1AA3 PROBLEM SET 2 – WEEK OF JANUARY 21, 2002

1.

What is the conjugate acid of each of the following species? (a) (b) (c) (d)

2.

Base S2− CH3COO− ClO2− SO32−

Conjugate acid HS− CH3COOH HClO2 HSO3−

What is the conjugate base of each of the following species? (a) (b) (c) (d)

3.

SOLUTIONS

Acid H2CO3 HBr H2PO4− [(CH3)2NH2]+ (dimethylammonium ion)

Conjugate base HCO3− Br− HPO42− (CH3)2NH2 (dimethylamine)

Write balanced chemical reactions and Ka or Kb expressions for the first ionization of the following acids and bases (i.e. assume in each case that only one H+ is ionized or added). (a) HF

(b) H3PO4

(c) CH3NH2

(d) CH3COOH

F− + H3O+

(e) HClO3

Ka = [F−][ H3O+] [HF]

(a)

HF + H2O

(b)

H3PO4 + H2O

H2PO4− + H3O+

Ka = [H2PO4−][ H3O+] [H3PO4]

(c)

CH3NH2 + H2O

CH3NH3+ + OH−

Kb = [CH3NH3+] [OH−] [CH3NH2]

(d)

CH3COOH + H2O

(e)

HClO3 is a strong acid and is fully dissociated at all concentrations in water. There is no equilibrium expression to describe the behaviour of this acid.

H3O+ + CH3COO−

Ka = [H3O+] [CH3COO−] [CH3COOH]

2 4.

(a) (b) (c) (d)

What is the pH of a 0.035 M solution of Ba(OH)2 (aq)? What is the pH of a 0.0042 M solution of HClO4 (aq)? What is the pOH of the solution in part (b)? A 30.0 mL sample of HClO4 was diluted to 500 mL, and the pH of the final solution was 4.56. What was the concentration of HClO4 in the original sample?

SOLUTION: (a)

Ba(OH)2 (aq) is a soluble hydroxide, and a strong base. It dissociates fully in water, so Ba(OH)2 (aq) produces Ba2+ (aq) and 2 −OH (aq). The original 0.035 M sample of Ba(OH)2(aq) therefore produces 0.035 M Ba2+ ions, and 2 × (0.035M) −OH ions (according to the formula of our hydroxide). ∴ [−OH]

=

2(0.035M)

From this we can find pOH: pOH

=

0.070 M

= = =

-log10[−OH] -log10[0.070] 1.15

From pH + pOH = 14, rearrange to get pH = 14 – pOH (b)

=

14 – 1.15

=

12.85 = 12.9

HClO4 is a strong acid, and dissociates fully in water to give: HClO4(aq) + H2O(l) → ClO4−(aq) + H3O+(aq)

The original 0.0042 M sample of HClO4(aq) therefore produces 0.0042 M ClO4− ions, 0.0042 M H3O+ ions

and

pH = -log10[H3O+]

=

-log10(0.0042) =

2.38

(c)

pOH = 14 – pOH

=

14 – 2.38

11.62 = 11.6

(d)

Since HClO4 is a strong acid, it dissociates fully in solution. In the final solution, pH = 4.56, and we can find [H3O+] directly from pH:

=

3 Since pH = −log10[H3O+], rearrange to give −pH = log10[H3O+] and take the antilog to give Thus, [H3O+] = 10−4.56

=

10−pH = [H3O+]

2.75 × 10−5 mol/L

To find the initial concentration of HClO4, we need the number of moles of HClO4 present in solution: mol HClO4

= =

(2.75 × 10−5 mol/L)(0.500 L) 1.38 × 10−5 mol

Original concentration of HClO4 = = 1.38 × 10−5 mol 0.0300 L 5.

(a) (a)

mol HClO4 original volume

= 4.60 × 10−4 M

What is the pH of a 10.0 M solution of HCl? What is the pH of a 1.00 × 10−8 M solution of HCl?

SOLUTION: (a)

pH = −log10[H3O+] = −log10(10.0) = −1 Very acidic! The point of this question is just to show that the pH scale doesn't just go from 0 to 14, it's just that most of the substances we talk about lie within that range.

(b)

Again,

pH

=

−log10[H3O+]

BUT in this case, what is [H3O+]? Remember that from the autoionization of water, there is a concentration of [H3O+] = 10−7 M (which we usually ignore in our calculations). In this case, however, the concentration is very significant. Thus, we have two sources of H3O+ in our solution: the H2O and the HCl: ∴[H3O+]

= =

Now we can use

1.00 ×10−8 M (from HCl) + 1.00 × 10−7 M (from water) 1.10 × 10−7 M pH = −log10[H3O+] = −log10(1.10 × 10−7) = 6.96

(If you ignore the [H3O+] from water in this case, the pH will work out to be 8, which doesn't make sense, since we're talking about a dilute solution of acid!).

4 6.

Dimethylamine, (CH3)2NH, is a weak base (ionization constant Kb = 7.40 × 10−4). (a)

What is the equilibrium concentration of dimethylammonium ion, [(CH3)2NH2]+ in a 0.400 M aqueous solution of (CH3)2NH?

(b)

What is the pH of a 0.400 M aqueous solution of (CH3)2NH?

SOLUTION: (a)

This is a “weak base” problem. Start by writing down the chemistry: (CH3)2NH

(CH3)2NH2+

+ H2O

+

−

OH

Set up an “ICE” table (Initial, Change, Equilibrium concentrations) in order to obtain mathematical expressions that describe the reaction. (CH3)2NH Initial conc. Change Equilibrium conc.

−

(CH3)2NH2+

0.400 M -x 0.400-x

OH

(10−7 M from water)* +x x

0 +x x

* We ignore the amount of −OH present from the ionization of water, if the value (10−7 M) is very small, as compared to x. Since (CH3)2NH functions as a base in our equilibrium, set up a Kb expression: Kb

=

[(CH3)2NH2+][ −OH] = [(CH3)2NH]

7.4 × 10−4

Substitute the equilibrium concentration values from the “ICE” table into Kb: (x)(x) (0.40-x)

=

7.4 × 10−4

Assume x