Chem104
Quiz 7 - Answers
Self Check Due by 5PM Wednesday.
1. Four species are combined in equimolar amounts: hydrocyanic acid, cyanide, acetate
and acetic acid. Write the balanced equilibrium reaction among these and say which
species will predominate (have the highest concentrations).
You need to write the equation with an acid and base on each side.
Iniitally it doesn’really matter which side has which species (in my opinion).
HCN + acetate <==> CN- + acetic acid
pKa of HCN is 9.40 while pKa of acetic acid 4.47, so acetic acid is the stronger acid.
Incidentally this also menas the pKb of CN- is 4.60 and acetate is 9.53, so cyanide is the stronger
base.
Therefore the equilibrium of the initial equimolar solution will shift in favor of the weaker acid
(and base). One could indicate the shift to the reagents (as written) as:
HCN + acetate <====----> CN- + acetic acid
2. Using the Ka value from lecture for phosphoric acid, calculate pKb2 and give the specific
equilibrium reaction that is associated with Kb2.
First you must figure out which equilibrium corresponds to pKb2.
Start with the most basic species in the phosphoric acid system: PO43-.
Kb1 corresponds to the equilibrium PO43- + H2O <==> HPO42- + OHKb2 corresponds to the equilibrium HPO42- + H2O <==> H2PO4- + OHKb3 corresponds to the equilibrium H2PO4- + H2O <==> H3PO4 + OHFor H2PO4-, Ka2 = 6.2 x 10-8 so pKa2 = 7.21.
pKb2 = 14- 7.21 = 6.79 (which corresponds to the Kb2 given on p.770)
3. You are doing so well I decided to give you the problem I decided to not do in class today.
Because it may be still a stretch fro some of you, you may email me with questions about it
and I'll provide hints.
5 mL of NaOH (0.100M) is added to 50 mL of acetic acid (0.100 M).
What is the pH?.
1. Write the reaction:
AcOH + NaOH  AcO- + Na+ + H2O where the arrow indicates this neutralization will
go to completion for as much NaOH is available. All the NaOH will react.
2. Do the stoichiometry:
# mol NaOH = # mol OH- = 0.005 L x 0.100M = 0.00050 mol OH# mol AcOH = 0.050 L x 0.100M = 0.0050 mol AcOH:
the acetic acid is in excess and all the NaOH will be consumed.
After the neutralization reaction there will remain:
AcOH
+ NaOH  AcO- + Na+ + H2O
0.0045 mol
0 mol
0.0005mol
Total volume is now 55mL so **initially** the concentrations are:
AcOH
+
0.0045 mol
0.055L
NaOH  AcO- + Na+ + H2O
0 mol
0.0005 mol
0.055L
0.0818M
0.0091 M
3. Re-establish equilibrium between remaining AcOH and formed AcO-:
The relevant equation is:
AcOH + H2O <==> AcO- + H3O+ described by a Ka where
Ka = [AcO-] [H3O+] / [AcOH] = 1.8 x 10-5
Set up ice box:
[AcOH]
[AcO-]
[H3O+]
Initial
Change
Equil
0.0818
-x
0.0818 -x
0.0091 M
+x
0.0091 + x
0M
+x
+x
Write the reaction:
AcOH + H2O <==> AcO- + H3O+ **remember the AcOH is still in excess**
Where
Ka = 1.8 x 10-5 = [AcO-] [H3O+] / [AcOH]
Make substitutions: 1.8 x 10-5 = (0.0091 + x)(x) / (0.0818 -x)
Now assume (0.0091 + x) ~ 0.0091 and assume (0.0818 -x) ~ 0.0818 , i.e., that x is very
small
Then 1.8 x 10-5 = (0.0091)(x) / (0.0818 )
x = 0.00016 M = [H3O+]
pH = -log 0.00016 = 3.79