Just as we used Ksp values to help us calculate ion concentrations in Unit 4: Solutions, we will use Ka and Kb to calculate the concentrations of ions in acids and bases. This will be crucial to determining pH later in this unit, so be sure to follow along carefully.
This is one of the most important sections of this unit mathematically, so be sure you understand every step.
How you calculate ion concentrations depends on whether you have a strong acid (or base) or a weak acid (or base). |
Calculating Ion Concentrations for Strong Acids & Bases
For strong acids and bases, the concentration of the ions can be readily calculated from the balanced equation. Consider these examples carefully.
1. Calculate the hydrogen ion concentration in a 0.050 M solution of hydrochloric acid. Solution:
We know that HCl is a strong acid that ionizes completely in water (you should memorize the list of strong acids). Begin by writing the balanced reaction:
HCl(aq) → H+(aq) + Cl-(aq)
From the balanced equation we see that 1 mole of HCl produces 1 mole of H+ (a 1:1 ratio), therefore the concentration of H+ will equal that of HCl.
Answer: [H+ ] = 0.050 M
2. Calculate the hydroxide ion concentration in a 0.010 M solution of barium hydroxide, Ba(OH)2. Barium hydroxide is a strong base. Solution:
Always begin by writing a balanced equation:
Ba(OH)2 (aq) → Ba2+(aq) + 2 OH-(aq)
Since 2 moles of OH- are produced for every 1 mole of Ba(OH)2 , the concentration of OH- will be twice the concentration of Ba(OH)2 .
Answer:
[OH-] = 2 × 0.010 = 0.020 MCalculating Ion Concentrations for Weak Acids & Bases
Weak acids and bases require a much different approach to finding ion concentrations. Once you know you have a weak acid or base, follow these steps in finding ion concentrations:
- Write a balanced equation for the reaction
- You will need to know the value of Ka or Kb - if it is not given in the question, look it up in a Table of Acid and Base Strengths.
- Set up the equilibrium constant expression. You will know the value of Ka (or Kb) and the concentration of the acid; you will be solving the equation for the concentration of the ions.
Follow along with these examples very carefully!
1. Calculate the hydrogen ion concentration in a 0.10 M acetic acid solution, HC2H3O2.
Ka for acetic acid, a weak acid, is 1.8 ×10-5.Solution:
Begin by writing the balanced reaction:
HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq)
The question gives us the concentration of the acid, HC2H3O2 (0.10 M).
We need to find the concentration of H+, which will also equal the concentration of C2H3O2- (why?)
Because ionization is NOT complete because this is a weak acid , [H+] will NOT equal [HC2H3O2]. Instead we must calculate it using the equilibrium constant expression.
Set up the
Ka equation:
[HC2H3O2]Substitute values into the equation. Let x equal the unknowns
0.10Rearrange the equation (1.8 ×10-5)(0.10) 1.8 ×10-6 Take the square root 1.3×10-3 ANSWER 1.3×10-3 1.3×10-3 An important note - in our Ka equation we used a value of 0.10M for [HC2H3O2]. This isn't quite correct. At equilibrium, the concentration of HC2H3O2 will actually be less:
initial [HC2H3O2] subtract the lost [H+] equilibrium [HC2H3O2] Using this equilibrium concentration of [HC2H3O2] in our calculations above (instead of 0.10 M), however, does not change our answer significantly. Therefore for our calculations we can safely ignore the negligible change in concentration of the acid.
2. Calculate the hydroxide ion concentration, [OH-], in a 0.025 M solution of analine, C6H5NH2, a weak base with Kb = 4.3×10-10 Solution:
This question will test some of your skills. Begin by writing a balanced equation. Since analine is a base that doesn't contain the hydroxide ion, include H2O as a reactant. Also remember that a base is a hydrogen acceptor and will gain an additional H+:
C6H5NH2 (aq) + H2O (l) ↔ C6H5NH3+(aq) + OH-(aq)
As we did in the previous example, we now set up the Kb expression and solve for ion concentrations. We see from the balanced equation that the ions have a 1:1 ratio, therefore [OH-] will equal the [C6H5NH3+].
Set up the Kb equation, omitting liquid water:
[C6H5NH2]Substitute values into the equation. Let x equal the unknowns
0.025Rearrange the equation (4.3×10-10)(0.25) 1.1×10-11 Take the square root 3.3×10-6 ANSWER 1.3×10-3 M 1.3×10-3 M Finding [OH-] in Acids and [H+] in Bases
Remember Kw from the previous section? Now we learn why it is important.
When we need to determine ion concentrations of an acid, you should immediately realize you will be finding the concentration of hydrogen ion (H+) and some anion (Cl- and C2H3O2- in our acid examples above). The concentration of the anion normally doesn't concern us much.When finding ion concentrations of bases, we determine the concentration of hydroxide ions, OH-, and some cation (Ba2+ and C6H5NH3+ in our base examples above). Again, we aren't too concerned with the cation concentrations.
Remember water?
H2O(l) → H+(aq) + OH-(aq)
When we make an acid solution, the hydrogen ion concentration will increase because we are adding more H+ ions to those already present in water.
Consider our strong acid example from above, in which [H+] = 0.05 M. If we add this to the hydrogen ion concentration of pure water, 1.0×10-7 (see last section), we get a total hydrogen ion concentration of 0.0500001M. Clearly, the water adds little to the total and we can essentially ignore its contribution.
[H+] from water: 1.0×10-7 [H+] from HCl Total [H+] 0.050 001 Next, remember that equilibrium constants are A CONSTANT (as long as temperature does not change). Thus the value of Kw will still have a value of 1.0 ×10-14 even though [H+] has increased due to the presence of the acid. We can use this information to calculate the concentration of hydroxide ions present in the aqueous solution:
Rearrange the equation
[H+]Substitute known values and solve for [OH- ]
0.05Le Châtalier's Principle
Remember Le Châtalier's Principle? When we disrupt an equilibrium system by increasing the concentration of a reaction participant, equilibrium will shift to minimize the stress. When we increase the H+ ion concentration in the water equilibrium system, the reaction will shift to the left to "use up" the additional H+. This will cause the concentration of OH- to decrease. Indeed, we see that [OH-] does decrease, from an original concentration of 1.0×10-7 in pure water to 2.0 ×10-13 in our acid solution.
We can apply the same calculations to determine hydrogen ion concentration in any basic solution. Let's return to our weak base, analine, example from above.
We determined that in a 0.025 M solution of analine, C6H5NH2, the [OH-] was 1.3 ×10-3 M. Our new question - what is the H+ ion concentration in this basic solution?
To solve, we again refer to our pure water equilibrium and its equilibrium constant expression. This time, however, we have determined [OH-] and need to find [ H+].
Again it should be apparent that the contribution to [OH-] from the water, 1.0×10-7 M, will have no significant effect on the OH- concentration so we can ignore it from our calculations:
[OH-] from water: 1.0×10-7 [OH-] from C6H5NH2 1.3 ×10-3 + Total [OH-] 0.001 3001 Use water's equilibrium constant to determine [H+]:
Rearrange the equation
[OH-]
Substitute in known values and calculate [H+]
1.3 ×10-3
For any acid or base
you can calculate
both [H+] and [OH-]Acids
First determine [H+]
then use Kw to calculate [OH-]Bases
First determine [OH-]
then use Kw to calculate [H+]
Some things to think about:
- In water, which is neutral (neither acidic nor basic), [H+] = 1.0×10-7M and [OH-] = 1.0×10-7M
- Acids increase [H+], so [H+] will be greater than 1.0×10-7 and [OH-] will be less than 1.0×10-7M
- Bases increase [OH-], so [OH-] will be greater than 1.0×10-7, and [H+] will be less than 1.0×10-7M
The calculations discussed in this section are very important to your understanding of acids and bases. Take time to work start Practice Set 2. Complete questions 1 - 7 now; you will finish this practice set after the next section.
