CHEM 101/105
Stoichiometry, as applied to Aqueous Solutions containing Ionic Solutes
Lect-05
MOLES - a quantity of substance. Quantities of substances can be expressed as masses, as numbers, or as
moles. It is equally informative to make reference to 2.578 grams of sulfur, 4.17 E-22 molecules of methane,
or 0.103 moles of fluorine. For the moment dwell on the mass and mole quantities.
ONE MOLE of all substances contains the same number of items. Turn this concept around a little bit. Given a
mass quantity of a substance how can the number of moles it represents be determined? The number of
moles in a given mass (weight in grams) is equal to that part of the standard mass it corresponds to.
The standard mass of antimony (at.no. 51) is 121.757 grams per mole. How many moles are contained in
7.4201 grams of antimony? (ANS: Compare the given mass of antimony (7.4201 g) to the standard mass
(121.757 g). What part of the standard mass is the given mass? Or, how many times does the standard mass
go into the given mass?, i.e., part / whole) Use the suggested method of setting up such conversions…
? moles Sb = 7.4201 g Sb
 1 mole Sb 
= 0.0609 moles Sb
 121757
.
g Sb 

(ARANF)
SOLUTIONS are homogeneous mixtures that consist of three parts:
(What is the third part?)
one part is the solute (component present in lesser amount), and
another part is the solvent (component present in greater amount).
In aqueous solutions the solvent is water.
The solution process Ionic bonding exists only in the crystalline solid state. Consequently, when ionic substances
are dissolved in water, ionic bonds are broken. This requires energy.
The fact that aqueous solutions of ionic substances are quite common suggests that the solution process
compensates for energy required to break ionic bonds. Part of this energy is provided by the free and independent
migration of oppositely charged ions in solution, which, in the solid, were locked in rigid positions. Additional energy is
provided b/c of stabilizing interactions between ions and water molecules. Water is a polar molecule, i.e., the molecule
has a non-uniform distribution of charge, although there is no net charge. The more positive regions on water
molecules (the H atoms) tend to orient towards anions in solution, and the more negative regions (the O atoms) tend
to orient towards cations in solution. This preferential orientation is the major driving force of the solution process for
ionic solutes and is represented in chemical formulas by appending a condition subscript of
(aq).
So
NaCl (s)
refers to solid sodium chloride, whereas NaCl (aq) refers to an aqueous solution containing the separated ions of
sodium chloride. (DEMO conductivity of electrolyte vs non-electrolyte aqueous solutions.)
1. Expressing CONCENTRATIONS of substances present in solutions
The unit of concentration of interest
at this time is the MOLARITY unit. Molarity is defined as a ratio of moles of solute over Volume of solution
expressed in Liter units. The symbol for molarity is capital M, and the math expression is...
MOLARITY = (moles solute) / (vol.solution in liter units)
A.
B.
C.
D.
Describe how to prepare:
M = moles / L
i.
500 mL of a 0.15 M solution of potassium iodide;
ii.
0.250 L of 0.085 M potassium dichromate.
How many moles are in:
i.
37.42 mL of a 0.127 M solution of sodium hypochlorite;
ii.
6.37 x 10 -2 L of 0.0832 M barium nitrate.
How many mL of a 0.0931 M solution of HCl will contain:
i.
5.00 x 10 -3 moles of hydrogen ion;
ii.
1.50 grams of HCl.
How many moles of ions are present in 43.29 mL of: i.
1.25 M barium hydroxide;
ii.
0.641 M chromium(III) chloride.
Notice that the product of (vol.solution in LITERS) x (conc. in MOLARITY) = moles solute, and this is the path
connecting solutions to all of the mass/number/mole relationships already met in Chapter 3. It will always be
true that any two substances (say A and B) in any reaction will be related as follows:
(moles of A) = (moles of B) x (conversion factor(s))
This relation between moles A and B provides the stoichiometric basis for every chemical reaction.
(moles of A) = (moles of B) x (conversion factor(s))
Examples of Solution Stoichiometry Problems
A.
A 0.1027 M solution of nickel(II) sulfate ( Ni SO 4 ) is reacted with a 0.0973 M solution of
lead(II) nitrate to form the insoluble substance lead(II) sulfate.
Write a balanced net ionic equation for the reaction.
What volume of the lead(II) nitrate solution is needed to react with 50.00 mL of
the nickel(II) sulfate solution?
B.
Hydrochloric acid ( H Cl ) reacts with calcium hydroxide [ Ca ( OH ) 2 ] in a "neutralization"
reaction ( reaction of acid with base forming a "salt" and water).
Write a balanced net ionic equation for the reaction.
In a one particular reaction 29.73 mL of the acid was needed to completely react with
25.00 mL of the base. If the concentration of the acid is known to be 0.153 M, what is the
concentration of the base?
C.
A 0.562 M solution of silver(I) nitrate ( Ag NO 3 ) is to be reacted with a 0.0631 M solution
of scandium(III) chloride ( Sc Cl 3 ) to form the insoluble substance silver(I) chloride
( Ag Cl ).
Write a balanced net ionic equation for the reaction.
What volume of the scandium chloride solution will be needed to react completely with 43.15
mL of the silver(I) nitrate solution?
D.
In order to produce complete reaction, 25.00 mL of a lanthanum chloride solution ( La Cl 3 )
are needed to react with 13.85 mL of a 0.0225 M solution of sodium oxalate ( Na 2 C 2 O 4 )
to form the insoluble substance lanthanum oxalate [ La 2 ( C 2 O 4 ) 3 ].
Write a balanced net ionic equation for the reaction.
What is the molarity of the lanthanum chloride solution?
What mass of precipitate would be formed?
Answers
1. Expressing CONCENTRATIONS of substances present in solutions
Molarity =
A.
i.
moles solute


 vol.solution in Liter units 
solve for moles of potassium iodide needed to form 500 mL of a 0.15 M solution.
moles KI = ( molarity ) x (liters solutions) = (0.15 M ) x ( 0.500 L ) = 0.075 moles KI
convert to grams KI:
directions:
ii.
? g KI = 0.075 moles KI
weigh out 12.45 g KI, and add enough water to completely dissolve
all solute, and to attain a final combined volume of 500 mL.
? moles K 2 Cr 2 O 7 = (0.085 M) x (0.250 L) = 0.02125 moles K 2 Cr 2 O 7
convert to grams: ? g K 2 Cr 2 O 7 = 0.02125 moles
directions:
B.
i.
ii.
C.
i.
 166g ( KI ) 
1mole( KI )  = 12.45 g KI


 294g 
1mole  = 6.2475 g
weigh out 6.2475 g K 2 Cr 2 O 7 , and add enough water to completely
dissolve all solute, and attain a final combined volume of 250 mL.
? moles Na Cl O = ( 0.127 M ) x ( 0.3742 L) = 0.04752
? mole Ba ( N O 3 ) 2 = ( 0.0832 M ) x (0.0637 L) = 0.00530
hydrochloric acid is a STRONG acid, i.e., it is completely ionized in solution.
Each HCl furnishes ONE hydrogen ion, so molarity HCl = molarity H 1+ (aq)
? L = ( 5.00 E - 3 moles H 1+ (aq) ) / ( 0.0931 M ) = 0.05371 Liters
or 53.71 mL
ii.
suggest converting 1.50 grams to moles HCl first, and then finding volume of solution
 1mol( HCl ) 
 = 0.0411
36
.
5
g
(
HCl
)


? moles HCl = 1.50 g HCl 
? L = ( 0.0411 moles HCl ) / ( 0.0931 M ) = 0.4414 L or 441.4 mL
D.
i.
Barium hydroxide, Ba ( OH ) 2 , is completely ionized in aqueous solutions, it is a
STRONG base. The balanced equation showing this process would be:
Ba ( OH ) 2
(s)
---> Ba 2+
(aq)
+ 2 OH 1-
(aq)
Note that THREE ions are provided by each formu la unit, so whatever the concentration of
Ba ( OH ) 2 , the TOTAL concentration of ions will be THREE times as great.
[ Ba ( OH ) 2 ] = 1.25 M
so
[ ions ] = 3.75 M
? moles ions = ( 3.75 M in ions ) x ( 0.04329 L ) = 0.162
ii.
Chromium(III) chloride, Cr Cl 3 , is a salt and will be completely ionized in aqueous solution.
The balanced equation showing this process would be:
Cr Cl 3
(s)
---> Cr 3+
(aq)
+ 3 Cl 1-
(aq)
Note that FOUR ions are provided by each formula, so whatever the concentration
of Cr Cl 3 , the TOTAL concentration of ions will be FOUR times as great.
[ Cr Cl 3 ] = 0.641 M
so
[ ions ] = 2.56 M
? moles ions = ( 2.56 M in ions ) x ( 0.04329 L ) = 0.111
worked-out Examples of Solution Stoichiometry Problems follow:
A. Starting Information:
Concentrations expressed
in Terms of...
0.1027 M
reactant
solution A
But what's really present in the solution is...
2+
Ni( aq
)
NiSO 4
SO42−(aq)
+
50 mL
and
0.0973 M
reactant
2+
Pb( aq
)
Pb(NO 3 ) 2
solution B
+
2 NO31(−aq )
? VOL
Will any combination above produce a net change?
Net Ionic equation:
Lead(II) cations and sulfate anions
will form insoluble lead(II) sulfate:
2+
Pb( aq
)
+
SO42(−aq)
→
PbSO4 ( s)
moles A = moles B [ C.F.]
moles Pb(NO 3 ) 2 = moles NiSO 4 [ C.F. ]
 1 mole SO42−  1 mole Pb2+  1 mole Pb( NO3) 2 
( 0.0973 M ) ( ? L) = ( 0.1027 M)(0.050 L) 

2−  
2+

1 mole NiSO4  1 mole SO4   1 mole Pb
? Vol. in mL = 52.77
B. Starting Information:
Concentrations expressed
in Terms of...
0.153 M
reactant
1+
H(aq)
HCL
solution A
But what's really present in the solution is...
1−
Cl(aq)
+
29.73 mL
and
?M
reactant
solution B
2+
Ca(aq)
Ca( OH ) 2
+
1−
2 OH (aq)
25.00 mL
Will any combination above produce a net change?
Net Ionic equation:
Hydrogen cations and hydroxide anions
will form water in a neutralization reaction:
1+
H(aq)
+
1−
OH(aq)
→ H2O (liq.)
moles A = moles B [ C.F.]
moles Ca( OH ) 2 = moles HCl [ C.F. ]
 1 mole H 1+   1 mole OH 1-   1 mole Ca(OH)2 
( ? M ) (25.00 mL) = ( 0.153 M)(29.73 mL) 

1+  
1- 
 1 mole HCl   1 mole H   2 mole OH 
? M = 0.091
C. Starting Information:
Concentrations expressed
in Terms of...
0.562 M
reactant
Ag(1aq+ )
AgNO 3
solution A
But what's really present in the solution is...
+
NO31−(aq)
43.15 mL
and
0.0631 M
reactant
3+
Sc(aq)
ScCl 3
solution B
1−
3 Cl (aq)
+
? VOL
Will any combination above produce a net change?
Net Ionic equation:
Silver(I) cations and chloride anions
will form insoluble silver(I) chloride:
1+
Ag(aq)
+
2−
Cl (aq)
→
AgCl (s)
moles A = moles B [ C.F.]
moles ScCl 3 = moles ScCl 3 [ C.F. ]
 1 mole Ag 1+  1 mole Cl1-  1 mole ScCl3 
( 0.0631 M ) ( ? mL) = ( 0.562 M)(43.15 mL) 

1+  
1− 
1 mole AgNO3 1 mole Ag   3 mole Cl 
? mL = 128.1
D. Starting Information:
Concentrations expressed
in Terms of...
? M
reactant
solution A
But what's really present in the solution is...
3+
La(aq)
LaCl 3
−
+ 3 Cl 1(aq)
25.00 mL
and
0.0225 M
reactant
solution B
Na 2 C 2 O 4
2 Na( 1aq+ )
+
C2O42(−aq)
13.85 mL
Will any combination above produce a net change?
Net Ionic equation:
Lanthanum(III) cations and oxalate anions
will form insoluble lanthanum(III) oxalate:
3+
2 La(aq)
+
−
3 C2 O42(aq)
→
La 2(C2 O4 )3(s)
moles A = moles B [ C.F.]
moles LaCl 3 = moles Na 2 C 2 O 4 [ C.F. ]
 1 mole C2 O42 −   2 mole La3 +  1 mole LaCl3 
( ? M ) ( 25.00 m L) = ( 0.0225 M)(13.85 mL) 

2−  
3+ 
 1 mole Na2 C2O4   3 mole C2 O4   1 mole La 
? M = 0.00831