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Equilibrium Intro

Disturbing Equilibrium

Disturbing Equil'm Answers

Pretest 3.1

Law of Chem Equil'm

Pretest 3.2

Acids/Bases: Intro

 

Ka/Kb:Notes,etc

Pretest 3.3

Student Pages(More coming!)

Redox: Oxidation Numbers



Dissociation Constants for Weak Acids

Exercises
Solutions
Solutions(Continued)

A weak acid is one with a reasonably small K, which implies that only a small percent of the acid is actually dissociating into H+ and a negative ion.

For HA(aq) = H+(aq) + A -1(aq)

In all the examples we will see there will only be one H+1 coming out of the acid at any one time, so the ratio of [HA]dissociating into [H+1] will be 1:1, and the ratio of [H+1] to [A -1] will be 1:1.

The above expression then becomes:

KA is an indicator of how strong an acid is compared to a different acid. If KA is very large there is 1 mole of H+ for every initial mole of HX. So a 0.10 M solution will have a pH of 1. If KA is low, the acid is said to be weak and an initial concentration identical to the above will lead to a higher pH.

Note the wording in the following:

 

Acid

KA

pH

 

HCl

very large

1.0

A concentrated solution of a strong acid(strong because of high Ka)

4.0

A diluted solution of a strong acid

H(CH3COO)

1.8 X 10-5

5.1

Diluted solution of a weak acid

2.4

More concentrated solution of a weak acid

 

 

Example 1: If Ka for the acid HA is 1.00 X 10-5, what is the pH of an acid with an initial concentration of 0.100 M?

HA(aq) = H+1(aq) + A -1(aq).

Let x = [H+1] and cross multiply:

x2 = 1.00 X 10-6 - 1.00 X 10-5x

x2 + 1.00 X 10-5 x - 1.00 X 10-6 = 0.

x = [H+1] = 9.95 X 10 -4.

pH = -log[H+1] = -log[9.95 X 10 -4] = 3.00

Example 2 A 0.100 mol/L solution of HCH3CO2 is partially ionized. At equilibrium [H+1] = 1.34 X 10-3.

Calculate the equilibrium constant of acetic acid, given:

HCH3CO2(aq)--> H+1(aq) + CH3CO2 -1(aq).

HCH3CO2(aq)

H+1(aq)

CH3CO2 -1(aq).

Initial

0.100

0

0

Reacting

1.34 X 10-3.

1.34 X 10-3.

1.34 X 10-3.

Equilibrium

0.100-1.34 X 10-3= 9.87 X 10-2 .

1.34 X 10-3.

1.34 X 10-3.

M

B. KB

 

KB = dissociation constant for a base. Handle all Kb problems as if they were KA problems, except that they relate to bases, not acids.

 

 

 

Example 1 If the equilibrium concentration of XOH is 0.10 M, and the pOH = 3.0, what is the KB of XOH?

 

(First write an equation)

 

 

XOH(aq) =

X+(aq)

OH-(aq)

I

 

0

0

C

 

1:1 ratio, so also 10-3.0

10-3.0

E

0.10

0+ 10-3.0 =10-3.0

10-pOH = 10-3.0

 

KB= 1.0 X 10-5

 

 

Example 2

 

A chemist dissolved 0.3 moles of cocaine (C17H21NO4) in a liter of solution and its pOHwas 2.35.

First write an equation to show how it acts as a Bronsted-Lowry base in water, and then find Kb.

 

Cocaine is a topical anesthetic that was used in eye and throat surgery in the 19th and early 20th centuries. But in the 1900s it gained popularity as a recreational drug. Its highly addictive psychological properties often destroy people both financially and emotionally.

 

 

Let C17H21NO4 = X

 

 

 

X(aq) + H2O(l) =

X+(aq)

OH-(aq)

I

0.3moles/1.0 L = 0.3 M

0

0

C

1:1 ratio,

so also 10-2.35

1:1 ratio,

so also 10-2.35

10-2.35

E

0.3 -10-2.35

0+ 10-2.35=10-2.35

10-pOH = 10-2.35

 

KB= 7 X 10-5

 

 

 

 

Example 3

A policeman brings you a white powder that might be morphine sulphate (molar mass 383.1 g/mole), a strong pain killer that is physiologically addictive. Its pKB is 6.10. If the pH of a 100.0 ml sample is 10.4, how many grams of the white powder should appear upon evaporating the water, if the sample is truly morphine sulfate?

 

C17H21NO7S(aq) + H2O(l) = C17H21NHO7S+(aq) + OH- aq)

More on morphine? Read:

http://www.emsb.qc.ca/laurenhill/science/morphine.html

 

 

pH = 10.4

pOH =14 10.4 = 3.6

 

pKb = 6.10

Kb = 10-6.10

 

 

C17H21NO7S(aq) + H2O(l)

C17H21NHO7S+(aq)

OH- aq)

I

x = y + 10-3.6

0

0

C

1:1 ratio,

so also 10-3.6

1:1 ratio,

so also 10-3.6

10-3.6

E

y

0+ 10-2.35=10-3.6

10-pOH = 10-3.6

 

KB= 10-6.10

 

 

Y = =0.0794 moles /L

 

 

0.0794 moles /L*0.100 L = 0.00794 moles

 

0.00794 moles* 383.1 g/mole = 3.0 g