Name:
Solutions and Solubility
Part I: Multiple Choice ( /4)
1. Which of the following is soluble in water?
a.
b.
c.
d.
Lead (II) iodide
Lithium sulfate
Silver chloride
Chromium (III) sulfide
2. To determine if nitrate is present in the local drinking water, what
cation could be added to the sample to see a precipitation reaction?
a.
b.
c.
d.
chloride
sulfate
nothing would precipitate
nitrate is already a solid silly
3. On a solubility graph, if an increase in temperature increases
solubility, it would look like:
a.
b.
c.
d.
a curve upwards
a curve downwards
a circle
a square
4. An evil wizard has asked you to choose which of the following an
example of a net ionic equation is. Be careful and good luck!!
a.
b.
c.
d.
AgNO3(aq) + KI(aq)  AgI (s) + KNO3 ( aq)
CaCl2  Ca 2+ + 2ClAg + (aq) ; Ba 2+(aq) ; Mg 2+(aq)
Ag + (aq) + Cl – (aq)  AgCl (s)
Part II: Short Answer
Compare and contrast the following terms: ( /6)
a) Solvent and Solute
1. Solute: The substance of lesser quantity that is dissolving inside
of the solute
2. Solvent: The substance of greater quantity that is dissolving the
solute
b) Dissociation and Ionization
1. Dissociation occurs when an ionic compound (consisting of a
metal [or NH4+ ion] and a non metal) dissolves into its ions in
water
2. Ionization occurs when an acid (consisting of an H+ ion and a
non metal) dissolves into its ions in water
Part III: Calculations
1. What volume in mL of 36.0M sulfuric acid is needed to contain 3.45g
of the acid? ( /3)
3.45g/(98g/mol)=.035 mol
L=Mmol L=(36M)(.035mol) L=1.26 1.26L=1260mL
2. What is the molarity of 8.95g of sodium carbonate in an 800mL
solution? ( /3)
8.95g/(102g/mol)=.0877 mol 800mL=.8L
M=.0877mol/.8L M=.109 .109M
3. What weight in grams of sodium hydroxide is there in 3.50L of a
0.60mol/L sodium hydroxide solution? ( /3)
M=mol/L mol=0.60Mx3.50L mol=2.1
2.1molx38g/mol=79.8g of NaOH
4. Calculate the mass in grams of solute in a 8.95ppm of phosphorus in a
soil sample with a mass of 2.2kg. ( /3) 2.2kg=2200g
8.95 ppm/1,000,000=solute/2200g solute=.01969 g of P
5. Calculate the mass in grams of a solute in a 10% sodium chloride
solution if the mass of the solvent is 350grams. ( /3)
10%/100=solute/350 solute=35g of NaCl
6. You need 250mL of a 0.40M NaCl solution but the only supply of NaCl
you have is a 6.0M solution of NaCl. What volume do you need ( /3)
and explain how to prepare it ( /4).
M1V1=M2V2 6.0M(V1)=(250mL)(.40M) V1=16.66mL
1. Add ~120mL water to the volumetric flask, 250mL in size
2. Add 16.66mL of the stock solution of NaCl
3. Add the remaining water to the volumetric flask, filling it up to
the line
4. Agitate
7. Using your solubility graph determine if the following are soluble or
insoluble. (/5)
a)
b)
c)
d)
e)
CuCl2 Soluble
Na2CO3 Soluble
Na2S Soluble
KI Soluble
Mg3(PO4)2 Insoluble
Part IV: Longer Answer
1. Write the total net ionic equations for the following reactions:
a) Aqueous barium hydroxide reacts with aqueous iron (III)
sulfate. ( /3)
3Ba+2(aq)+ 6OH- (aq)+ 2Fe+3(aq)+3SO4-2(aq) 3BaSO4 (s) + 2Fe(OH)3 (s)
b) Aqueous barium bromide reacts with aqueous sodium sulfate.
( /3)
Ba+2(aq)+SO4+2(aq)  BaSO4(s)
c) CuSO4 (aq) + NaOH (aq) Cu(OH)2(s) + Na2SO4 (aq)
2. Use the following solubility graph to answer the following questions.
a) Which substance is most soluble at 60 C? Explain. ( /2)
i. Potassium Iodide, because it has the highest solubility of
any other substance on the graph.
b) Which two substances have the same solubility at 80 C? ( /2)
i. Potassium Chlorate
ii. Sodium Chloride
c) Which substances solubility is most dependant on temperature?
Explain.(/2)
i. Potassium nitrate, because it’s solubility changes the
most over the temperature curve.
d) At what temperature does potassium iodide have a solubility of
150g/100mL water? ( /1)
i. ~13 C
e) You have a solution of potassium chlorate containing 4g at 65 C.
How many more grams would you have to add to make this a
saturated solution? Explain. ( /2)
i. Add 26g extra, because that would put 30 g in the solution
which is the point where (at 65C) it becomes saturated.