CHEM 101
RECALL:
Stoichiometry, as applied to Aqueous Solutions containing Ionic Solutes
Lect-04
ONE MOLE contains 6.02 x 10 23 items, and has a mass (in grams) equal to its Standard Mass.
A fractional part of a mole is that same fractional part of 6.02 x 10 23, and also of the Standard Mass.
This concept is the source of countless conversion factors involving any pairing of three factors –
moles,
numbers
and masses.
A chemical formula shows the (1) number and (2) type of atoms present in a substance.
Subscripts in chemical formulas represent moles of the atoms they reference.
Coefficients in a balanced chemical equation represent moles of the substances they reference.
SOLUTIONS are homogeneous mixtures that consist of three parts:
one part is the solute (component present in lesser amount), and
another part is the solvent (component present in greater amount).
(What is the third part?)
In aqueous solutions the solvent is water.
The solution process
Ionic bonding exists only in the crystalline solid state.
Consequently, when ionic substances dissolve in water, ionic bonds in the crystal are broken.
This requires energy.
The fact that aqueous solutions containing ionic solutes are common suggests the solution process is able to
compensate 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 (an entropy effect). Additional
energy is provided by stabilizing interactions between ions and water molecules (an enthalpy effect).
Water is a polar molecule, i.e., the molecule has a non-uniform distribution of charge, even though it is electrically neutral
overall. More positive regions on water molecules (the H atoms) tend to orient towards solute anions in solution, and more
negative regions (the O atoms) tend to orient towards solute cations in solution. This preferential orientation is the major
driving force for the solution process for ionic solutes. (Show CD-ROM Videos. - ppm 4.1, and DEMO conductivity of
electrolyte vs non-electrolyte aqueous solutions.)
1. Expressing CONCENTRATIONS of substances present in solutions
The unit of concentration of interest here is
MOLARITY. Molarity is defined as moles solute per Liter of solution. The symbol for molarity is capital M, and
the math expression is...
MOLARITY = (moles solute) / (vol.solution in liter units)
or
M = moles / L
A.
Describe how to prepare:
i.
ii.
500 mL of a 0.15 M solution of potassium iodide;
0.250 L of 0.085 M potassium dichromate.
B.
How many moles are in:
i.
ii.
37.42 mL of a 0.127 M solution of sodium hypochlorite;
6.37 x 10-2 L of 0.0832 M barium nitrate.
C.
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.
D.
How many moles of IONS are present in 43.29 mL of:
i.
ii.
1.25 M barium hydroxide;
0.641 M chromium(III) chloride.
Notice the product of (vol.solution in LITERS) x (conc. in MOLARITY) = moles solute, and this is the path connecting
solutions to all mass/number/mole relationships already met in Chapter 3. Furthermore, it is universally true that quantities
of any two substances (A and B for example) in any reaction will be always related by the expression:
(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.
Ionic solutes are involved in three types of reactions in aqueous solutions:
precipitation reactions - sparingly soluble ionic substances are produced when solutions are mixed
acid/base reactions - involves hydrogen and hydroxide ions (the constituent ions of water)
oxidation/reduction reactions - involves gain and loss (i.e., exchange) of electrons between reactants
2. Precipitation reactions in aqueous solutions
What happens when two solutions with ionic solutes are mixed?
ANS: A precipitate may, or may not form.
A decision must be made about whether some combination of ions present in the mixed solutions can form a sparingly soluble
substance, i.e., whether or not a precipitate could form, by testing or from previous experience.
DEMO this by testing solutions of nickel(II) nitrate, barium chloride, sodium carbonate, copper(II) sulfate, and potassium
hydroxide. In each case:
(a) Start by listing the ions present in the separated solutions.
(b) Test whether any combination of ions forms an insoluble substance upon mixing.
(c) If a precipitate forms then write a net ionic equation to represent the process.
1.
When a solution of sodium carbonate is mixed with one of barium chloride does a precipitate form?
Do the test. If yes, then either one or both of the ionic salts sodium chloride, and/or barium carbonate is insoluble.
A ppt does form, and it is BaCO3. Write the net ionic equation for this precipitation reaction:
Ba2+ (aq) + CO32-(aq) = BaCO3(s)
Note that sodium ions and chloride ions do not appear in the net ionic equation, because there is no net change in their
condition; they begin as aquated Na1+ (aq) and Cl1-(aq) ions, and remain as aquated Na1+ (aq) and Cl1-(aq) ions both before and after the precipitation reaction. So they are not included in the net ionic equation. They are called
"spectator" ions, they go along for the ride but do not participate in precipitation formation.
2.
Mix solutions of nickel(II) nitrate and potassium hydroxide.
Does a precipitate form?
Do the test. If yes, then some combination of the ions present in solution must have formed an insoluble ionic substance,
i.e., a precipitate. Lets analyze these solutions, decide what precipitated, and write a net ionic balanced equation for the
precipitation reaction.
3.
Mix solutions of copper(II) sulfate and potassium hydroxide.
Does a precipitate form?
Do the test. If yes, then some combination of the ions present in solution must have formed an insoluble ionic substance,
i.e., a precipitate. Analyze starting solutions and decide what precipitated. Write a net ionic balanced equation for the
precipitation reaction.
Results of such tests and previous experience lead to Rules of Solubility, used to predict whether or not an insoluble
product will be formed. Solubility Rules are presented in terms of ions forming insoluble substances. Combinations of ions
shown in the table below form precipitates, i.e., substances that are sparingly soluble in water.
IONS
Gp I & NH 4 1 +
Gp II Cations
T.M. & Other Cations
NO 3 1 -
Cl 1 -
Ag
1+
, Hg 2
2+
, Pb
SO 4 2 -
OH 1 -
CO 3 2 -
PO 4 3 -
S 2-
Ba 2 + , Sr 2 + , Pb 2 +
Mg 2 +
ALL
ALL
ALL
ALL
ALL
ALL
2+
Solubility rules are grouped and listed according to anions as follows:
If nitrate anion is present (NO3-1), the salt is soluble. This is worded as, "all NITRATES are soluble".
Most salts containing chloride anion (Cl-1) are soluble except when the cation is Ag+1, Hg2+2, or Pb+2.
This is abbreviated to, "all chlorides are soluble, except AgCl, Hg2Cl2, and PbCl2".
Most salts containing sulfate anion (SO4-2) are soluble,
except BaSO4, SrSO4, and PbSO4.
Most salts containing carbonate anion (CO3-2), or hydroxide anion (OH-1) are insoluble,
except GROUP I and NH4 1+ cations, and as noted.
Most salts containing sulfide anions (S -2) are insoluble,
except main GROUPS I and II cations, and ammonium cation (NH4+1).
Apply these "solubility rules" to write balanced net ionic chemical equations for each of the examples below. In addition,
solve any associated stoichiometric questions.
A.
Will a precipitate form when solutions of aluminum sulfate and sodium chloride are mixed?
B.
Will a precipitate form when solutions of ammonium chloride and lead(II) nitrate are mixed?
(How many grams precipitate could be formed from reaction of 42.78 mL of 0.125 M NH4Cl ?)
C.
Will a precipitate form when solutions of uranium(III) nitrate and sodium carbonate are mixed?
(How many grams precipitate can be formed from reaction of 29.31 mL of 0.0951 M Na2CO3 ,
with 32.13 mL of 0.0544 M U(NO 3) 3 ? FW of U2(CO3)3 = 656 amu )
D.
It is desired to remove iron(II) ions from a water supply. What anion(s) might be used?
3. Aqueous acid/base reactions
a. a little bit more about water
Water is a polar substance. This means water is able to "solvate" ions rather well.
Another important characteristic of water is its ability to undergo self-ionization into
hydrogen and hydroxide ions - to a very slight degree, as shown by the equation
HOH = H+1(aq) + OH-1(aq).
At a given temperature the amount of water ionized (i.e., present as hydrogen ions and hydroxide ions) is very small but
remains constant.
This causes the product of hydrogen ion concentration and hydroxide ion concentration to be constant.
Due to the self-ionization water can be considered as a weak electrolyte.
The acidic, basic, or neutral condition of an aqueous solution depends on the relative concentrations of hydrogen and
hydroxide ions present. Aqueous solutions are neutral when [H+1] = [OH-1] , acidic when [H+1] > [OH-1] , and basic when
[H+1] < [OH-1]. However, no matter what the concentrations of these two ions, their product is constant.
The acidic/basic/neutral condition of aqueous solutions is often detected using small amounts of organic acids that display
colors ("indicators"). The color displayed under acidic conditions is different from the color under basic conditions.
b. acid/base chemistry - definitions and degree of ionization
DEFINED:
Acids donate protons or otherwise increase [ H1+ (aq) ] of the solution.
Bases accept protons or otherwise increase [ OH1- (aq) ] of the solution.
i. STRONG ACIDS
by
in aqueous solutions, are COMPLETELY dissociated into ions. For example hydrochloric acid,
having the formula HCl, exists in solution ONLY as the separated and independent ions - as shown
+
1−
HCl ( aq ) → H (1aq
) + Cl ( aq ) (note the ONE WAY arrow). NO HCl molecules are present in solution. Other
common strong acids are nitric, HNO3 ; sulfuric, H2SO4 ; perchloric, HClO4; hydrobromic, HBr ; and hydroiodic, HI
ii. WEAK ACIDS
in aqueous solutions, are only slightly dissociated into ions. For example, acetic acid,
having the formula, HC2H3O2 , exists primarily in molecular form, with few hydrogen and
acetate ions present as shown by
+
1−
HC2 H3 O2 (aq ) ↔ H (1aq
) + C2 H3 O ( aq ) (note the TWO WAY arrow ).
Based on its
very limited ionization, HC2H3O2 is called a WEAK acid. Other common weak acids are hydrofluoric, HF, formic,
HCOOH ; citric; butyric, i.e., all organic acids.
iii. STRONG BASES
In a parallel manner, strong bases are COMPLETELY ionized in aqueous solution. Examples are
hydroxides of group I alkali metals - MOH ; and group II alkaline earth - M(OH)2 .
iv. WEAK BASES
Weak bases exist largely in molecular form in aqueous solution, with very few ions present.
Examples are ammonia, NH3 ; and amines, such as CH3-NH2 . Weak bases operate in a manner slightly different from
weak acids; they do not furnish hydroxide ions directly, rather they accept protons (from solvent water molecules) and leave
hydroxide ions in their wake - as shown by
NH3 (aq) + HOH = NH4 1+(aq)
+ OH(aq) 1(note both charges and mass balance)
DEMO Test strong/weak acids/bases to show presence/absence of ions in aqueous solutions. Strong acids/bases are
completely ionized (bright light) Weak acids/bases are very slightly ionized (dim light).
Illustration of understanding about strong/weak acid/base character, requires writing the correct ionization equation for each
of the above cases as shown. For this purpose remember the above examples of the four substances: HCl, NaOH, NH3,
and HC2H3O2 and use them as models for strong acid and base, and weak base and acid respectively.
3. Acid / Base chemistry - stoichiometry When concentration of one component (the acid or base) is known, then it can be
used to determine the concentration of the other component by measuring
volumes required for complete reaction, and then using (moles A) = (moles B) x [Conversion Factor]. Burets and pipets
measure reacting volumes, and indicators signal the complete end-of-reaction. Such reaction processes are
called titrations.
Write a balanced net ionic equation for each of the following acid-base reactions in water:
A.
B.
C.
butyric acid (HC4H7O2, a weak acid) with potassium hydroxide (KOH, a strong base)
ammonia (NH3, a weak base) with hydrobromic acid (HBr, a strong acid)
nitric acid (HNO3, a strong acid) with lithium hydroxide (LiOH, a strong base)
Work each of the following acid / base stoichiometry problems:
D.
A titration required 32.78 mL of a 0.164 M potassium hydroxide solution, to neutralize 25.00 mL of a nitric
acid solution. What is the molarity of the nitric acid solution?
E.
Analysis of a barium hydroxide solution was conducted by titrating against hydrochloric acid. It was found
that 27.48 mL of a 0.215 M hydrochloric acid solution was required to neutralize 20.00 mL of a barium
hydroxide solution of unknown concentration. What is the molarity of the barium hydroxide solution?
Assume all hydroxide anions furnished by barium hydroxide are neutralized by hydrochloric acid.
F.
What volume of a 0.137 M solution of phosphoric acid, H3PO4 , would be required to neutralize 50.00 mL of
a 0.314 M solution of calcium hydroxide? H3PO4 is a polyprotic acid. Assume all hydroxide anions
furnished by calcium hydroxide and all protons furnished by phosphoric acid are involved in the neutralization
reaction.
G.
When 20.00 mL of a 0.154 M solution of hydrobromic acid, HBr , is slowly added to 45.00 mL of a 0.0941 M
solution of lithium hydroxide, LiOH, what is the final concentration of hydroxide ion in solution?
H.
Suppose 31.89 mL of a 0.114 M solution of sulfuric acid, H2SO4 , is slowly added to 26.21 mL of a 0.363 M
solution of sodium hydroxide, NaOH . What is the final concentration of the ion from water present in
excess? Assume all protons furnished by sulfuric acid are neutralized by sodium hydroxide.
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:
 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 = ( Molarity ) x ( Vol.Solution in Liters )
B.
C.
 166g ( KI ) 
1mole( KI )  = 12.45 g KI


i.
? moles Na Cl O = ( 0.127 M ) x ( 0.3742 L) = 0.04752
ii.
? mole Ba ( N O 3 ) 2 = ( 0.0832 M ) x (0.0637 L) = 0.00530
i.
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
ii.
or 53.71 mL
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
D.
i.
or 441.4 mL
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 formula 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
2. Precipitation reactions in aqueous solutions
A.
REACTANTS
IN SOLUTION
Al 2 (SO 4 ) 3
Al 3+
NaCl
Na 1+
(aq)
(aq)
&
SO 4 2-
&
Cl 1-
(aq)
(aq)
Address the question: Does any cation/anion combination result in a precipitate?
Check solubility rules, which show that no combination of ions results in an insoluble substance. So no
precipitate forms when these two solutions are mixed, and no net ionic equation is necessary in this case.
B.
REACTANTS
IN SOLUTION
NH 4 Cl
NH 4
1+
(aq)
Pb 2+
Pb ( NO 3 ) 2
(aq)
&
Cl 1-
&
NO 3 1-
(aq)
(aq)
Address the question: Do any cation/anion combinations result in a precipitate?
Check solubility rules, which show that insoluble lead(II) chloride would form.
The net ionic equation for the precipitation reaction is:
Pb 2+
i.
ii.
(aq)
+
2
Cl 1-
(aq)
=
Pb Cl 2
(s)
How many grams of precipitate can be formed from reaction
of 42.78 mL of 0.125 M NH 4 Cl with a solution of lead(II) nitrate?
Two different approaches to the solution of this problem will be illustrated:
(a)
using moles A = moles B [ C.F.'s …]
moles PbCl 2 ppt. = moles NH 4 Cl [ C.F.'s…]
recall:
for solutions,
for solids,
moles = M x Vol (in Liters)
moles = mass / Std.Mass
 mass PbCl2 

 = ( MNH4 Cl)(Vol. Liters)[ C. F.' s...]


FW
 ? mass 

 = (0.125M )( 0.04278 L)[C. F .' s...]
 278 
 1moleCl1−  1molePb2+  1molePb( NO3 ) 2 
[ C. F.' s...] = 1moleNH Cl 
1−  
2+

4

  2moleCl   1molePb
1  1  1
? grams PbCl 2 = (278)(0.125)(0.04278)
1  2  1 = 0.743 grams PbCl 2
(b)
using the precipitation equation and millimoles
DEFINE MILLIMOLES as the product of MOLARITY and VOLUME in MILLILITERS
Consider the precipitation reaction in three separate stages,
(1) before the reaction starts (B4),
(2) the actual reaction stage (Rx), and
(3) after the reaction is over (AFTR), and place in a MILLIMOLE GRID, as shown
Pb 2+
(aq)
+
2
Cl 1-
(aq)
--->
Pb Cl 2
(s)
B4
RX
AFTR
First step: determine the amounts of starting reactants in millimoles, before the reaction occurs:
? mmole Pb 2 + is present before the reaction, but
neither concentration nor volume of the lead(II) nitrate solution is specified by the
problem, so assume this reagent is present in an excess amount (LOTS).
? mmole Cl 1 - is present before the reaction and quantities given, determine the amount...
? mmole Cl 1 ? mmole Cl
1-
= ( M )x( mL ) [ C.F.'s…]
 1mmoleCl 1− 
= ( 0.125 M)( 42.78 mL) 

1mmoleNH 4Cl 
= 5.348
? mmole PbCl 2 is not present before the reaction occurs.
Enter these starting millimole amounts of reactants in the B4 line of the ppt. reaction:
B4
RX
AFTR
Pb 2+ (aq) +
LOTS
2
Cl 1- (aq) --->
5.348
Pb Cl 2 (s)
none
Second step involves activating the reaction according to the coefficients of the balanced ppt.
equation: i.e., ONE mmole Pb 2 + reacts with TWO mmole Cl 1 - for form ONE mmole PbCl 2 …
So ALL of the chloride anion will be used up in the reaction, along with half that millimole quantity
of Pb(II) cation. Reactants lose material so they carry negative millimole quantities. The ppt. is
formed so it carries a positive millimole quantity.
Pb 2+ (aq) +
LOTS
−2.674
B4
RX
AFTR
2
Cl 1- (aq) --->
5.348
−5.348
Pb Cl 2 (s)
none
2.674
Final step in this analysis involves subtracting reacting mmoles from starting mmoles,
to determine the mmoles remaining after the reaction is complete…
Pb 2+ (aq) +
LOTS
−2.674
LOTS
B4
RX
AFTR
2
Cl 1- (aq) --->
5.348
−5.348
none
Pb Cl 2 (s)
none
2.674
2.674
The millimole grid shows 2.674 mmoles of ppt has been formed.
However, the problem asks for this amount to be expressed in grams, so…
? g PbCl 2 = 2.674 x 10 - 3 moles PbCl 2
C.
REACTANTS
 278g 
1molePbCl  = 0.743 g PbCl 2

2
IN SOLUTION
U ( NO 3 ) 3
U 3+
Na 2 CO 3
Na 1+
(aq)
NO 3 1-
&
(aq)
&
CO 3 2-
(aq)
(aq)
Address the question: Does any cation/anion combination result in a precipitate?
Check solubility rules, which show uranium(III) carbonate could be formed as an insoluble substance.
A net ionic equation for the precipitation reaction would be:
U 3+
i.
ii.
(aq)
+
CO 3 2-
(aq)
=
U 2 ( CO 3 ) 3
(s)
Recognize that mole quantities of both reactants are provided by the problem, b/c volume and
molarity of both reactants are given (mole = M x L ). Consequently, trial calculations of the
quantity of ppt. formed from each reactant are needed in order to identify the limiting reagent.
Let's work things out in millimoles...
a.
trial calculation I
? mmole ppt. can be obtained from the uranium nitrate?
mmoles ppt = mmoles U ( NO 3 ) 3 [ C. F. ]
mmoles ppt = ( 0.0544 M ) ( 32.13 mL )
b.
trial calculation II
 1 mole U 3+   1 mole U2 (CO3 )3 


3+
 = 0.874 mmoles ppt.
 1 mole U ( NO3 ) 3   2 mole U
? mmole ppt. can be obtained from the sodium carbonate?
mmoles ppt = mmoles Na 2 CO 3 [C.F.]
 1 mole CO32-   1 mole U2(CO3 )3 
mmoles ppt = ( 0.0951) (29.39 mL ) 

 = 0.929 mmoles ppt.
2 1 mole Na2 CO3   3 mole CO3 
c.
U ( NO 3 ) 3 is the limiting reagent so base quantity of ppt. formed on it...
? g ppt. = 0.873 x 10 −3 moles
D.

656 grams 
 1 mole U (CO )  = 0.573 g U 2 (CO 3 )

2
3 3
3
ppt.
Seek an anion to form a precipitate with Fe(II) ions, but since this involves a water
supply source (for consumption) then the anion should not have any toxic or otherwise
objectionable features.
Sulfide anion will form insoluble FeS, but is objectionable for odor and other reasons.
Hydroxide anion will form insoluble Fe(OH) 2 , but requires an alkaline pH.
Of the set of ions in our solubility rules, carbonate is probably the best choice, but it too
has some objectionable qualities. (choices, choices, choices, …)
3.
Acid/Base chemistry - stoichiometry
A.
Butyric acid is an organic acid, and so it will be classified as a WEAK ACID.
Accordingly, it is but slightly ionized in aqueous solutions, most will be present
in the molecular acid form:
B4 solution
AFTR soln
H C 4 H 7 O 2 (aq)
lots
lots
H 1+ (aq) +
C 4 H 7 O 21 -(aq)
none
extremely small amounts
Potassium hydroxide is a STRONG BASE, and it will be completely ionized in aqueous solution:
B4 solution
AFTR solution
REACTION
KOH (s)
lots
none
K 1+ (aq) + OH 1 -(aq)
none
lots
Hydroxide anion (from KOH) will combine with the hydrogen of butyric acid to form water,
leaving butyrate anion, C 4 H 7 O 21 -(aq)
NET IONIC EQUATION
H C 4 H 7 O 2 (aq) + OH 1 -(aq) = HOH + C 4 H 7 O 21 -(aq)
[Notice that both sides carry a charge of same magnitude ( 1 ) and sign (negative), and that
potassium cation does not appear in the net ionic equation.]
B.
Hydrobromic acid is a STRONG ACID, and it will be completely ionized in aqueous solution:
B4 solution
AFTR solution
HBr (g)
lots
none
H 1+ (aq) + Br 1 -(aq)
none
lots
Ammonia is a WEAK BASE. It does not undergo self-ionization but requires water to produce ions, as
shown below. Nevertheless, very few ions are formed so most will be present in the molecular form.
B4 solution
AFTR solution
REACTION
NH 3 (g) + HOH
lots
lots
lots
lots
NH 4 1+ (aq) + OH 1 -(aq)
none
extremely small amounts
Hydrogen cation (from HBr) will combine with ammonia ( NH 3 )
to form ammonium cation ( NH 4 1+ )
NH 3 (aq) + H 1+-(aq) = NH 4 1+-(aq)
NET IONIC EQUATION
[Notice that both sides carry a charge of same magnitude ( 1 ) and sign (positive), and that
neither hydroxide nor bromide anions appear in the net ionic equation.]
C.
Nitric acid is a STRONG ACID, and it will be completely ionized in aqueous solution:
B4 solution
AFTR solution
H 1+ (aq)
none
lots
HNO 3 (ag)
lots
none
+ NO 3 1 -(aq)
Lithium hydroxide is a STRONG BASE, and it will be completely ionized in aqueous solution:
B4 solution
AFTR solution
LiOH (s)
lots
none
Li 1+ (aq) + OH 1 -(aq)
none
lots
REACTION
Hydrogen cation (from nitric acid) will combine with hydroxide anion (from LiOH)
to form molecular water.
EQUATION
H 1+ (aq)
+ OH 1 -(aq) = HOH
[Notice that both sides are electrically neutral, and that lithium cations and nitrate anions
do not appear in the net ionic equation.]
These three problems (A – C) illustrate and explain every type of STRONG/WEAK ACID/BASE interaction met in
CHEM 101. In problems that follow, only net ionic equations will be displayed. Refer to (A. - C.) for details.
D.
REACTION
KOH and HNO 3 are both STRONG electrolytes and are completely ionized in aqueous
solutions (like problem iii. above). Hydrogen cation (from nitric acid) will combine
with hydroxide anion (from KOH) to form molecular water.
EQUATION
H 1+ (aq)
+ OH 1 -(aq) = HOH
[Notice that both sides are electrically neutral, and that potassium cations and nitrate anions
do not appear in the net ionic equation.]
STOICHIOMETRY
moles HNO 3 = moles KOH [ C. F. ]
 1mole(OH1− )   1mole( H1+ )   1mole( HNO3 ) 
(? M ) (0.025 L) = (0.164 M)(0.03278 L) 


1−  
1+
 1mole(KOH)   1mole(OH )   1mole( H ) 
First CF via complete ionization of the STRONG BASE, KOH
Second CF via reaction of hydrogen and hydroxide ion to form molecular water
Third CF via complete ionization of the STRONG ACID, NHO 3
? M (HNO 3 ) = 0.215
E.
REACTION
Barium hydroxide and hydrochloric acid are both STRONG electrolytes and are
completely ionized in aqueous solutions (like problem iii. above). However, note that
each mole of Ba(OH) 2 will furnish TWO moles of hydroxide anions. Hydrogen cation
(from HCl) will combine with hydroxide anion (from Ba(OH) 2 ) to form molecular water.
EQUATION
H 1+ (aq)
+ OH 1 -(aq) = HOH
[Notice that both sides are electrically neutral, and that barium cations and chloride anions
do not appear in the net ionic equation.]
STOICHIOMETRY
moles Ba(OH) 2 = moles HCl [ C. F. ]
 1mole( H1+ )   1mole(OH1− )   1mole( Ba(OH)2 ) 
(? M ) (0.020 L) = (0.215 M)(0.02748 L) 


1+  
1−
 1mole( HCl)   1mole(H )   2mole(OH ) 
First CF via complete ionization of the STRONG ACID, HCl
Second CF via reaction of hydrogen and hydroxide ion to form molecular water
Third CF via complete ionization of the STRONG BASE, Ba(OH) 2
? M (Ba(OH) 2 ) = 0.148
F.
REACTION
Calcium hydroxide is a STRONG BASE and is completely ionized in aqueous solutions.
Phosphoric acid is a WEAK ACID and is but slightly ionized in aqueous solutions.
This neutralization reaction is like problem i. above). However, note that each mole of
H 3 PO 4 contains THREE moles of hydrogen cations, and that Ca(OH) 2 furnishes
TWO hydroxide anions. The hydroxide anions react with hydrogen ions to form water in
the neutralization reactionb.
H 1+
NEUTRALIZATION
STOICHIOMETRY
(aq)
+
OH 1 -(aq) = HOH
moles H 2 PO 4 = moles Ca(OH) 2 [ C. F. ]
 2 mole(OH1− )   1 mole H1+   1 mole( H3 PO4 ) 


1−  
1+
 1 mole(Ca(OH)2 )   1 mole OH   3 mole( H ) 
(0.137 M ) (? L) = (0.314 M)(0.050 L ) 
First CF via complete ionization of the STRONG BASE, Ca(OH) 2
Second CF via neutralization reaction.
Third CF via phosphoric acid furnishing three hydrogen ions.
? L (H 3 PO 4 ) = 0.0.0764 L
G.
or
76.40 mL
This problem is a bit different from the acid/base reactions met so far. Note that a volume of solution and
concentration of solution are given for both reactants, i.e., starting mole quantities are know for both
reactants. Recognize that this is a limiting reagent type of a problem; one of the reactants is present in
an excess amount and some of it will remain unreacted.
REACTION
LiOH and HBr are both STRONG electrolytes and are completely ionized in aqueous
solutions (like problem iii. above). Hydrogen cation (from hydrobromic acid) will combine
with hydroxide anion (from LiOH) to form molecular water. However, one of the
reactants is present in an excess amount and some of it will remain unreacted.
EQUATION
H 1+ (aq)
+ OH 1 -(aq) = HOH
[Notice that both sides are electrically neutral, and that lithium cations and bromide anions
do not appear in the net ionic equation.]
STOICHIOMETRY
Suggest a slightly different approach for limiting reagent types of problems
involving solutions. FIRST determine the moles of reacting ions as furnished
by the reactants.
?moles (H 1+ (aq) ) = moles (HBr) [ C. F. ]
 1mole( H1+ ) 
= (0.154 M)(0.020 L) 
 = 0.00308 moles H 1+(aq)
 1mole( HBr ) 
?moles (OH 1-(aq) ) = moles (LiOH) [ C. F. ]
 1mole(OH1− ) 
 = 0.00424 moles OH 1-(aq)
1
mole
(
LiOH
)


= (0.0941 M)(0.045 L) 
SECOND Rewrite the net ionic equation for the reaction and enter in a MOLE GRID.
Enter information in rows B4, REACT, and AFTR, as shown:
B4
REACT
AFTR
Note
H 1+ (aq) +
0.00308
-0.00308
none
OH 1 -(aq) =
0.00424
-0.00308
0.00116
HOH
lots
+0.00308
lots+0.00308
(1) initial moles quantities are entered in the B4 row.
(2) mole quantities in the REACT row MUST be in the same proportions as coefficients
in the net ionic equation (i.e., 1:1:1 in this case). Reactant quantities are negative
(i.e., used up) and product quantities are positive (i.e., formed).
(3) subtraction of REACT quantities from B4 quantities show which reactant is
completely used-up (i.e., the limiting reagent) and which one has some left after
reaction (i.e., the reagent present in excess)
THIRD Recall that 20.00 mL of one solution was added to 45.00 mL of the other solution, so the
total volume is now 65.00 mL. There are 0.00116 moles of Hydroxide anion remaining after
reaction. Its concentration would be:
 0.00116 mole 
 = 0.0178 M

0.065 L 
[ OH 1 -] = 
H.
This problem is similar to the above (vii.). It is another limiting reagent type of problem. The same
procedure used before will be applied here, with one change; we introduce and use units of millimoles.
REACTION
NaOH and H 2 SO 4 are both STRONG electrolytes and are completely ionized in
aqueous solutions (like problem iii. above). Hydrogen cation (from sulfuric acid) will
combine with hydroxide anion (from NaOH) to form molecular water. However, one of
the reactants is present in an excess amount and some of it will remain unreacted, AND
one mole of sulfuric acid furnishes TWO moles of hydrogen cation. The hydrogen and
hydroxide ions react to form molecular water.
EQUATION
H 1+ (aq)
+ OH 1 -(aq) = HOH
[Notice that both sides are electrically neutral, and that sodium cations and sulfate anions
do not appear in the net ionic equation.]
STOICHIOMETRY
Suggest a slightly different approach for limiting reagent types of problems
involving solutions. FIRST determine the moles of reacting ions as furnished
by the reactants. (Note that the product of MOLAR concentration and volume
in MILLILITER units yields a useful unit called MILLIMOLES.)
?moles (H 1+ (aq) ) = moles (H 2 SO 4 ) [ C. F. ]
 2mole( H1+ ) 
= (0.114 M)(31.89 mL) 
 = 7.17 millimoles H 1+(aq)
 1mole( H2 SO4 ) 
?moles (OH 1-(aq) ) = moles (NaOH) [ C. F. ]
 1mole( OH1− ) 
 = 9.51 millimoles OH 1-(aq)
1
mole
(
NaOH
)


= (0.363 M)(26.21 mL) 
SECOND Rewrite the net ionic equation for the reaction and enter in a MILLIMOLE GRID.
Enter information in rows B4, REACT, and AFTR, as shown:
B4
REACT
AFTR
Note
H 1+ (aq) +
7.17
-7.17
none
OH 1 -(aq) =
9.51
-7.17
2.24
HOH
lots
+7.17
lots+7.17
(1) initial millimoles quantities are entered in the B4 row.
(2) millimole quantities in the REACT row MUST be in the same proportions as
coefficients in the net ionic equation (i.e., 1:1:1 in this case). Reactant quantities
are negative (i.e., used up) and product quantities are positive (i.e., formed).
(3) subtraction of REACT quantities from B4 quantities show which reactant is
completely used-up (i.e., the limiting reagent) and which one has some left after
reaction (i.e., the reagent present in excess)
THIRD Recall that 31.89 mL of one solution was added to 26.21 mL of the other solution, so the
total volume is now 58.10 mL. There are 2.24 millimoles of Hydroxide anion remaining after
reaction. Its concentration would be:
 2.24 mmole 
[ OH 1 -] =  58.10 mL  = 0.0.0386 M
( MOLARITY = millimoles/ milliliters )