Kindy help me calculate Q1, Q2, Q3 and Q4. Buffers and How to Think About Them S
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Kindy help me calculate Q1, Q2, Q3 and Q4.
Buffers and How to Think About Them
So far, we’ve seen that we can calculate the pH of a solution of an acid or base, and we can calculate the pH of a solution of an acidic or basic salt. Suppose we mix two solutions: one of a weak acid, the other of a salt of the weak acid’s conjugate base. Each solution has concentrations of the weak acid and the conjugate base, and these depend on different equilibrium constants. When mixed, a new equilibrium must be established and we will have made a buffer. Buffers are cool and special because they resist large changes in pH when small amounts of strong acid or base are added. We’ll talk about how on Monday, but for now: How can we find the pH of the new buffer?
Let’s consider a model.
HA (aq) H+(aq) + A-(aq)
If we make a buffer, we will have some of the acid and some of the conjugate base in the solution. And since we made the solution, we know the concentrations of these ions initially. If we want to know the pH of the buffer, we need to find [H+] from these concentrations. To get this, let’s solve the Ka expression for [H+].
Ka for this reaction is:
Solved for [H+]:
If we take the –log of both sides, we will have a solution for pH:
The –log of Ka is pKa:
We are beginning to see a special relationship between the pKa of a weak acid and the pH of a buffer of that acid and conjugate base. Let’s rearrange the last term in the equation to form the Henderson-Hasselbalch equation:
If we know the pKa of the weak acid in our buffer (and we will, because we chose the buffer system) and the concentrations of the weak acid and conjugate base (which we will, because we chose to make the buffer at a certain concentration), we can solve for the expected pH of the buffer.
But before we do that, I’d like to point out some things about the Henderson-Hasselbalch equation.
If we make a buffer with equal amounts of the acid and conjugate base, the ratio will be 1:1 and the log of that will be 0. The pH will equal the pKa. This is the basis for our pKa of indicators lab! As the pH of the surrounding solution changes, so will the concentrations of the acid and base forms of the indicators. We can see the color change, so we know the pH at which [HA] = [A-] for each indicator.
Our experience with the indicators hints at another property of buffers. Like indicators, buffer systems have a useful range, usually taken to one pH unit above and below the pKa of the acid in the buffer. If we need a buffer to have a certain pH, we need to choose an acid that has a pKa near that pH for our buffer.
Because the acid and base concentrations are a ratio, we can see that multiple concentrations of a buffer may yield the same pH. The more concentrated the buffer is though, the better its buffering capacity.
Okay, enough words. Let’s do calculations! We’ll start with the acid and salt solutions prior to mixing. The final solutions are on the next page. I want you to show me the work needed to get to those solutions.
Q1. What is the pH of a 0.200 M HOCl (bleach) solution? Ka = 2.95 * 10-8
Q2. What is the pH of a 0.100 M NaOCl solution? (Find Kb first).
Q3. Suppose we mix these two solutions. Use the Henderson-Hasselbalch equation to find the pH of the buffer. (Note 1: Find pKa first. Note 2: The concentration of each ion is halved when we mix the solutions).
Q1. pH = 4.115
Q2. pH = 10.265
Q3. pH = 7.229
Let’s say we want to make a buffer. How would we do it?
Example: We want to make a buffer of pH = 5.00, and we want its concentration to be about 0.50 M. What should be our acid/base system, and how much of each will we need?
We need an acid with pKa 5.00. Acetic acid has a pKa of 4.76, so it would be a good option. Let’s put [HAc] = 0.50 M into the Henderson-Hasselbalch equation to see what our concentration of sodium acetate should be.
5.00 = 4.76 + log([Ac-]/[0.50 M])
0.24 = log([Ac-]/[0.50 M])
Taking both sides as a power of 10 gives:
1.738 = [Ac-]/[0.50 M]
[Ac-] = 0.87 M (two sig fig)
If our acetic acid concentration is 0.50 M, the acetate concentration must be 0.87 M to produce a pH = 5.00 buffer. The concentration of the base is greater than that of the acid, but as the pH is greater than the pKa of the acid, this seems reasonable.
Q4. Suppose you want to make an acetic acid/acetate buffer of pH = 4.60. If the buffer acid is approximately 0.30 M, what should the acetate concentration be?
Explanation / Answer
Q1 ) concentration = C = 0.200 M
Ka = 2.95 x 10^-8
[H+] = sqrt (Ka x C) = sqrt ( 2.95 x 10^-8 x 0.200) = 7.68 x 10^-5 M
pH = -log [H+] = -log (7.68 x 10^-5) = 4.11
pH = 4.115
Q2 )
Ka = 2.95 x 10^-8
pKa = -log Ka = 7.53
NaOCl is the salt of weak acid and strong base . so pH >7
pH = 7 + 1/2 [pKa + logC]
pH = 7 + 1/2 [7.53 + log 0.100]
pH = 10.265
Q3 )
pH = pKa + log [NaOCl / HOCl]
pH = 7.53 + log (0.1 / 0.20)
pH = 7.229
Q4 )
pH = pKa + log [acetate / acetic acid]
4.60 = 4.76 + log (acetate / 0.30)
acetate = 0.208 M
acetate ion concentration = 0.208 M
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