Chemistry 121
Worksheet 9 Notes
1.
Oregon State University
Dr. Richard L Nafshun
A student obtains 1.00 liter of an aqueous solution (a water solution) of 1.00 M HCl and
another 1.00 liter aqueous solution of 1.00 M propanoic acid. Which solution contains
the most H+ ions? Explain.
The 1.00 M HCl solution contains more H+ ions. HCl is a strong acid:
HCl → H+ + Cl-
HCl dissociates 100%; so there is no HCl present, it is all in the
form of H+ and Cl-.
Propanoic acid is a weak acid (it dissociates to a very small degree (less than 2%). There
is a considerable amount of CH3CH2COOH present and very little CH3CH2COO- and H+.
2.
Which of the following are soluble in water? NaCl, BaCl2, Ca(NO3)2, ammonium
carbonate, sodium phosphate, calcium phosphate. Explain.
NaCl is soluble—Rule 1 states that all sodium salts are soluble. This is a good exam
question.
BaCl2 is... I don't know... Our simplified solubility rules are no help. This is NOT a
good exam question.
Ca(NO3)2 is soluble—Rule 1 states that all nitrate salts are soluble. This is a good exam
question.
Ammonium carbonate is soluble—Rule 1 states that all ammonium salts are soluble.
This is a good exam question.
Sodium phosphate is soluble—Rule 1 states that all sodium salts are soluble. This is a
good exam question.
Calcium phosphate is insoluble—Rule 2 states that all phosphate salts are insoluble. This
is a good exam question.
3.
A student obtains 50.00 grams of calcium hydroxide. Determine the number of
hydroxide ions present in the sample. Determine the numbers of calcium ions present.
 1mol 
 0.6748 mol Ca(OH)2
50.00 grams Ca(OH)2 
 74.10 g 
 6.022 x10 23 Ca(OH ) 2 units 
 4.063 x 1023 Ca(OH)2 units
0.6748 mol Ca(OH)2 
1
mol


4.
 2OH  ions
4.063 x 1023 Ca(OH)2 units 
 1Ca(OH) 2 unit

 8.127 x 1023 OH- ions

 1Ca 2 ions
4.063 x 1023 Ca(OH)2 units 
 1Ca(OH) 2 unit

 4.063 x 1023 Ca2+ ions

Consider two aqueous solutions: one of sodium phosphate and one of calcium nitrate.
List the ions present in each. Is a precipitate formed when the two solutions are mixed?
If so, write the net ionic equation.
Full equation:
2 Na3PO4 (aq) + 3 Ca(NO3)2 (aq) → Ca3(PO4)2 (s) + 6 NaNO3 (aq)
Net ionic equation:
2 PO43- (aq) + 3 Ca2+ (aq) → Ca3(PO4)2 (s)
(Spectator ions do not appear in the net ionic equation. The spectator ions are Na+ and
NO3-).
5.
Consider two aqueous solutions: one of lithium hydroxide and one of calcium chloride.
List the ions present in each. Is a precipitate formed when the two solutions are mixed?
If so, write the net ionic equation.
Full equation:
2 LiOH (aq) + CaCl2 (aq) → Ca(OH)2 (s) + 2 LiCl (aq)
Net ionic equation:
2 OH- (aq) + Ca2+ (aq) → Ca(OH)2 (s)
(Spectator ions do not appear in the net ionic equation.
The spectator ions are Li+ and Cl-).
6.
Consider two aqueous solutions: one of ammonium carbonate and one of barium nitrate.
List the ions present in each. Is a precipitate formed when the two solutions are mixed?
If so, write the net ionic equation.
Full equation:
(NH4)2CO3 (aq) + Ba(NO3)2 (aq) → BaCO3 (s) + 2 NH4NO3 (aq)
Net ionic equation:
CO32- (aq) + Ba2+ (aq) → BaCO3 (s)
(Spectator ions do not appear in the net ionic equation. The spectator ions are NH4+ and
NO3-).
7.
Considering questions 4, 5, and 6 above, you propose a reaction and write the equation in
which two aqueous solutions produce a precipitate. Treat the two aqueous solutions as
though they are in separate beakers before they are mixed together in a larger beaker.
List the ions present in the beakers before and after the reaction. Write the net ionic
equation. Identify the spectator ions.
Many; how about (NH4)2S (aq) + Sr(NO3)2 (aq)?
8.
Why is ammonia (NH3) basic if it doesn’t contain hydroxide? Write a balanced equation
for the reaction of ammonia with water.
Ammonia is a weak base (not every ammonia molecule present in solution will accept a
proton, only a small percentage—hence a weak base). It accepts a proton:
NH3 (aq) + H+ (aq) ↔ NH4+ (aq)
9.
Write a balanced equation for the reaction of sodium hydroxide with hydrochloric acid.
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
10.
Write a balanced equation for the reaction of lithium hydroxide with nitric acid.
LiOH (aq) + HNO3 (aq) → LiNO3 (aq) + H2O (l)
11.
Describe the laboratory procedure known as a titration. What is a buret?
The laboratory procedure known as a titration is used to determine the concentration or
identify of an acid or a base. Consider the balanced equation for the reaction of sodium
hydroxide and hydrochloric acid:
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
If the concentration of one of the two reactants (sodium hydroxide or hydrochloric acid)
were not known, a comparison can be made. For example, you are given an unknown
HCl (aq) sample and titrate it with known HaOH (aq) (see Question 5 below).
12.
A student obtains 25.00 mL of an HCl solution of unknown concentration. Upon
titration, 21.35 mL of 0.1200 M NaOH are required for neutralization. Determine the
concentration of the HCl solution.
The reaction: NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
The apparatus:
The buret (cylindrical glassware) is filled with a known concentration of NaOH (aq);
0.1200 M NaOH (aq). The Erlenmeyer flask below contains the sample of HCl (aq);
25.00 mL of unknown concentration. The NaOH (aq) solution is drained from the buret
into the flask where it mixes with the HCl (aq) to form salt water:
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
When the moles of NaOH added equals the moles of HCl (aq) in the sample, the flask
contains NaCl (aq) and water. No HCl (aq) remains and no excess NaOH (aq) was
added. The reaction is complete. At this point (known as the equivalence point):
molesNaOH (aq) = molesHCl (aq)
MNaOHVNaOH = MHClVHCl
The above equation (MNaOHVNaOH = MHClVHCl) is true because M is the concentration in
moles/L and V is the volume in L:
MV = (mol/L)(L) = moles
MNaOHVNaOH = MHClVHCl
(0.1200 M)( 0.02135 L) = (MHCl)(0.02500 L)
MHCl =
13.
(A)
(0.1200 M)( 0.02135 L)
= 0.1025 M
(0.02500 L)
What is meant by a strong acid? Name and write chemical formulae for three
strong acids. Write the dissociation reactions for these strong acids.
A strong acid produces protons (H+) and it dissociates 100% to produce a lot of protons.
HCl (aq) → H+ (aq) + Cl- (aq)
(Dissociation of hydrochloric acid)
HNO3 (aq) → H+ (aq) + NO3- (aq)
(Dissociation of nitric acid)
H2SO4 (aq) → H+ (aq) + HSO4- (aq)
(Dissociation of sulfuric acid)
(B)
What is meant by a weak acid? Name and write the chemical formula of a weak
acid. Write the dissociation reaction for this weak acid.
A weak acid produces protons (H+) and it dissociates less than 100% to produce a few
protons. The –COOH group (the carboxylic acid group) is a weak acid group.
CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq) (Dissociation of acetic acid. Acetic acid
dissociates to a very small degree—only one in every 2000 acetic acid molecules
dissociates to produce protons and this piece: CH3COO- (aq))
(C)
What is meant by a strong base? Name and write chemical formulae for three
strong bases. Write the dissociation reactions for these strong bases.
A strong base accepts protons (H+) and it accepts all available protons. Hydroxide ion
(OH-) acts as a strong base.
NaOH is completely soluble:
NaOH (aq) → Na+ (aq) + OH- (aq)
will do some chemistry)
OH- (aq) + H+ (aq) → H2O (l)
(Sodium ion is a spectator ion and hydroxide
NaOH, LiOH, and KOH are strong bases.
(D)
What is meant by a weak base? Name and write the chemical formula of a weak
base. Write the dissociation reaction for this weak acid.
A weak base accepts protons (H+) and it few available protons. Ammonia (NH3) acts as a
weak base.
NH3 (aq) + H+ (aq) → NH4+ (aq)
14.
Identify a soluble carbonate salt. Identify an insoluble carbonate salt. Explain.
Na2CO3 is soluble (all sodium salts are soluble—Rule 1).
CaCO3 is insoluble (carbonates are insoluble—except those listed in Rule 1 and calcium
is not listed in Rule 1).
15.
Write a balanced chemical equation for a reaction that produces water and a salt.
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
This is the neutralization of a strong base and a strong acid.
16.
Write a balanced chemical equation for a reaction that produces carbon dioxide gas and
water (think carbonic acid).
[This is not important to the Chemistry 121 student at this time]
H2CO3 → H2O (l) + CO2 (g)
17.
A student titrates 0.290 grams of KHP (potassium hydrogen phthalate; MW=204.2
g/mol) to the equivalence point with 0.09950 M NaOH (aq). The volume of NaOH
solution required is:
NaOH (aq) + KHP (aq) → NaKP (aq) + H2O (l)
When the moles of NaOH added equals the moles of KHP (aq) in the sample, the flask
contains salt and water. No KHP (aq) remains and no excess NaOH (aq) was added. The
reaction is complete. At this point (known as the equivalence point):
molesNaOH (aq) = molesKHP (s)
Because KHP is weighed out (mass and molar mass were provided) leave the right side
of the equation to enter moles.
MNaOHVNaOH = molesKHP (s)
0.290 grams of KHP
204.2 g/mol
VNaOH = 0.01427 L or 14.27 mL
(0.09950 M)(VNaOH) =
(A)
(B)
(C)
(D)
(E)
70.08 mL.
35.04 mL.
31.45 mL.
140.16 mL.
14.27 mL.