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
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.]
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:
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:
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:
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 = 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:
Using our data, this gives
1.00 ´ 10–3 mole = 0.200 mole/L ´
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
= 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
= 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
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:
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.
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 0.40M HCl solution. The equivalence point is shown using dotted lines, with another line pointing to where the curve intersects with the equivalence point.
– 5 – Raynor: 9/23/2010