GENERAL CHEMISTRY
SOLVING STRONG ACID – STRONG BASE PROBLEMS
I.
PRIMARY SUBSTANCE PRESENT IS A STRONG ACID OR A STRONG BASE:
No equilibrium. Determine [H3O+] and/or [OH–] directly from the chemical reaction equation and
the concentrations of acid or base present.
Example 1:
What are the concentrations of all ions in a 0.40 M solution of Ba(OH)2?
Ba(OH)2 is a strong soluble base. Thus it is 100% ionized. The balanced reaction is
H2O
Ba(OH)2 + 
 Ba2+ (aq) + 2 OH– (aq).
final [Ba2+] = 1  0.40 = 0.40 M
final [OH–] = 2  0.40 = 0.80 M
II. STRONG ACID – STRONG BASE TITRATIONS:
As we titrate, we are slowly adding a base to an acid solution or an acid to a base solution, so the
volume in the receiving beaker is increasing during the titration. That means the concentrations of
all its components are also changing. So, when doing titration calculations, it is very important to
keep track of both the numbers of moles of all components as well as the total volume in the
container. Do not calculate the concentrations until you have determined how many moles of each
substance are present and the total volume in the container.
The equivalence point is defined as follows:
# moles acid = # moles base
To determine the relationship, we must use a balanced chemical reaction equation. Alternatively, we
can note that we always have the same net ionic equation when we react strong acid with strong
soluble base. This net ionic equation is given by
H+ (aq) + OH– (aq)  H2O (l)
This leads to the following relationship:
# moles of H+ in solution = # moles of OH–.
Whether or not the solution is acidic or basic will depend upon which substance is in excess. If
more H+ remains after the reaction is complete, the solutions will be acidic, but if more OH–
remains after the reaction is complete, the solutions will be basic.
[Remember: moles A = [A]  Vtot.]
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Example 2:
A titration begins with 10.0 mL of 0.200 M HCl in a beaker. A solution of 0.200 M Ba(OH)2 is
added slowly until the equivalence point is reached. How many mL of base were needed to
completely neutralize the acid?
First, (as always), calculate the # moles of any substances you can! We have a known
concentration and volume for the HCl solution. We can therefore determine the # moles HCl:
Moles HCl = MHCl  VHCl (in Liters) = 0.200 mole/L  10.0 mL  10–3L/1 mL
Moles HCl = 2.00  10–3 mole
Note, if you use the net ionic equation, (Method B below), you will need to convert moles acid
into moles H+, using the chemical formula:
Moles H   Moles HCl 
1 Mole H 
 Moles HCl
1 Mole HCl
Thus, for this problem, Moles H+ = 2.00  10–3 moles.
Next, write the balanced chemical reaction equation for the reaction taking place. [In either its
full form or in net ionic form, depending on which way is easier for you. I will show both
below as Method A and Method B.]
We will now determine how many moles of Ba(OH)2 are needed, using either the full balanced
chemical reaction equation , (Method A), or the net ionic equation, (Method B):
Method A – Using the full (balanced!) chemical reaction equation:
The balanced chemical reaction equation is given below:
Ba(OH)2 (aq)+ 2 HCl (aq)  BaCl2 (aq) + 2 H2O (l)
At the equivalence point, (i.e., when all the acid or base is neutralized), we must have the
same number of moles acid as moles base. We know how many moles of acid are present,
so we know that we must convert the moles of HCl into moles of Ba(OH)2, using the
appropriate balancing coefficients:
Moles Ba(OH) 2  Moles HCl 
1 Mole Ba(OH) 2 1
  Moles HCl
2 Moles HCl
2
Thus, we get Moles Ba(OH)2 = ½ (2.00  10–3 mole ) = 1.00  10–3 mole
Method B – Using the (balanced!) net ionic equation:
The balanced chemical reaction equation is given below:
H+ (aq) + OH– (aq)  H2O (l)
At the equivalence point, (i.e., when all the acid or base is neutralized), we must have the
same number of moles H+ as moles OH–. We know how many moles of acid are present,
so we know that we must convert the moles of HCl into moles of Ba(OH)2, using the
appropriate balancing coefficients:
Moles OH   Moles H + 
1 Mole OH 
 Moles H +
1 Mole H +
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Thus, we get Moles OH– = 2.00  10–3 mole
In order to calculate the volume of Ba(OH)2 solution needed from its molarity and moles,
we must first determine the relationship between moles OH– and moles Ba(OH)2, using
the chemical formula:
Moles Ba(OH) 2  Moles OH  
1 Mole Ba(OH) 2 1
 Moles OH 

2 Moles OH
2
Moles Ba(OH)2 = ½ (2.00  10–3 mole ) = 1.00  10–3 mole
Both methods yield the total number of moles of Ba(OH)2 needed to neutralize all of the acid.
We can now use this information, along with the molarity of the base solution, (given to us), to
determine the volume needed:
Moles Ba(OH) 2  M Ba(OH)2  VBa(OH)2 (in Liters)
Using our data, this gives
1.00  10–3 mole = 0.200 mole/L  VBa(OH)2
We can now solve for the volume, (in Liters), by dividing both sides by the molarity:
V = 1.00  10–3 mole / 0.200 mole/L = 5.00  10–3 L
Finally, we can convert the volume into mL using 1 mL = 10–3 L. The final volume in mL is
VBa(OH)2 = 5.00 mL.
Example 3:
What are the concentrations of each ion in solution after 40.0 mL of 0.200 M HCl is added to
30.0 mL of 0.200 M Ba(OH)2.
As always, our first step is to calculate the number of moles present of each substance we have
data for. We have concentrations and molarities of both HCl and Ba(OH)2, so we are able to
calculate how many moles of each of them will be present.
Moles HCl = MHCl  VHCl (in Liters) = 0.200 mole/L  40.0 mL  10–3L/1 mL
Moles HCl = 8.00  10–3 mole
Moles Ba(OH) 2  M Ba(OH)2  VBa(OH)2 = 0.200 mole/L  30.0 mL  10–3L/1 mL
Moles Ba(OH)2 = 6.00  10–3 mole
Next, write down the balanced chemical reaction that occurs. Remember, the net ionic equation
has the same form when we add a strong soluble acid to a strong soluble base:
H+ (aq) + OH– (aq)  H2O (l)
Note that the concentrations of Ba2+ and of Cl– that are in the solution will remain unchanged
during the reaction – they are both spectator ions. It is only the H+ and OH– ions that are
changed during the reaction. So lets calculate the number of moles of all these ions that we start
with, (by using the chemical formulas of the acid and base):
For HCl, we see that moles H+ = moles Cl– = moles HCl. Thus, we get
Initial moles H+ = 8.00  10–3 mole
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Initial moles Cl– = Final moles Cl– = 8.00  10–3 mole
For Ba(OH)2, we see that moles Ba+2 = moles Ba(OH)2 and moles OH– = 2  moles Ba(OH)2.
We get
Initial moles Ba+2 = Final moles Ba+2 = 6.00  10–3 mole
Initial moles OH– = 2  6.00  10–3 mole = 1.20  10–2 mole
To determine the final moles of H+ and OH– we note that there is a 1 to 1 relationship between
them in the net ionic equation. Thus whichever has the greater number of moles will be in
excess and the other will be the limiting reagent. For the data given above, this means that the
OH– is the limiting reagent, (and will be completely used up during the reaction), and the H+
will be in excess. We now know that the solution will be acidic. We now know the following
result:
Final moles H+ = 0  [H+] = 0.
To determine moles OH– remaining in excess, we first have to determine how much of it
reacts. We can then subtract that amount from the total amount we had initially:
Moles OH– reacting = moles H+ reacting = 8.00  10–3 mole
Moles OH– remaining in solution = Initial moles OH– – moles OH– reacted
= 1.20  10–2 initial moles – 8.00  10–3 moles reacted
Moles OH– remaining in solution = 4.0  10–3 moles
Finally, to determine the concentrations of each ion in solution, we have to divide by the total
volume. Since we added two solutions together, it will be given by the sum of the volumes:
VTOT = Vacid + Vbase = 40.0 mL + 30.0 mL = 70.0 mL
For molarities, we must use the volume in Liters. Converting, we get
VTOT = 7.00  10–2 L
The final concentrations for each ion are as follows:
6.00 103 moles
 Ba 2  
 8.6 102 M
2
7.0 10 L
8.00 103 moles
Cl  
 0.11 M
7.0 102 L
0 moles
 H   
0 M
7.0 102 L
4.00 103 moles
OH   
 5.7 102 M
7.0 102 L
III. REDOX TITRATIONS:
Redox titrations work the same way as acid-base titrations with only one difference – we must use
the balanced chemical reaction equation for the redox reaction, instead of one for acids with bases.
We carry out the calculations in exactly the same manner as was discussed in the examples above,
using the coefficients for the balanced redox reaction instead of those for an acid-base reaction, to
convert moles of one substance in the reaction into moles of another.
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To find the equivalence point, do the following: suppose we are titrating a solution of known
volume and molarity for substance A and to it we are slowly adding a solution containing substance
B. The equivalence point is the point at which the number of moles of B that have been added are
the amount needed to completely use up substance A. [Think of substance A as being the limiting
reagent.] You need to use the balancing coefficients to convert the known moles of substance A into
moles of substance B – in exactly the same manner as that used in Example 2, (where HCl was
substance A and Ba(OH)2 was substance B). Then, if you are asked for the volume needed of
solution B, use its known molarity and moles to determine its volume, (in the same manner as in
Example 2).
If you need to calculate final molarities of any of the substances after two solutions were added
together, (as was done in the Example 3), think of it as a straight-forward limiting reagent problem.
Remember that the limiting reagent always determines the moles of all reactants that get used up
and also determines the moles of all the products, and is itself completely used up. [Don’t forget –
you must use the balancing coefficients to convert moles of the limiting reagent into moles of
another substance in the chemical reaction.] Then, to calculate final molarities, you need to know
the final volume. It will be equal to the sum of the volumes of the two solutions that were mixed
together. [See Example 3]
IV. TITRATION CURVES:
A strong acid – strong base titration curve is shown on the last page. The drawing shown is for a
titration where a 0.20 M OH– solution is slowly added to 100.0 mL of a 0.40 M HCl solution. The
equivalence point is shown using dotted lines, with another line pointing to where the curve
intersects with the equivalence point.
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