The pH of human blood needs to be between 7.35 and 7.45. You want to prepare a buffer
solution that gives a pH of 7.40. You decide to use a sodium phosphate buffer: the acid is
H2PO−
4 and the conjugate base is HPO2−
4
. You want the concentration of the acid to be
0.0100 M.
1. If the initial H2PO−
4
concentration is 0.0100 M, what is the initial concentration of
HPO2−
4
that will give a pH of 7.40?
2. What is the maximum molarity of acid that this buffer can neutralize without the pH
dropping below 7.35?
3. What is the maximum molarity of base that this buffer can neutralize without the pH
going above 7.45?
I keep trying to post this question but i can;'t find it once i post it.. i was help with part one, but i cannot figure out part 2 and 3
You might help yourself if you used a name that isn't used often. Anonymous is used by several people so you can't tell which is your post.
I responded last night that I didn't think there was enough information to answer part 2 and part 3. You MUST hav a volume in that question somewhere. A simple M depends upon HOW MUCH of that M solution you have.
I have responded several times that I don't think it can be done without know the volume. I've changed my mind. I believe I know that they are asking for now. Here is what you do.
The problem you have for 2 gives you a 7.35 value. You know the base and acid for 7.40. Plug in 7.35 for the pH into the HH equation and the base you have, and calculate the M acid.
For 3 you use 7.45 and the acid you have and calculate the M of base you can add from the HH equation. Post you work if you get stuck.
To solve these problems, we will use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the conjugate acid and the conjugate base.
1. For part one, we are given the initial concentration of the acid (H2PO−
4) as 0.0100 M. We want to find the initial concentration of the conjugate base (HPO2−
4) that will give a pH of 7.40. The Henderson-Hasselbalch equation is:
pH = pKa + log([base]/[acid])
In this case, the acid is H2PO−
4 and the conjugate base is HPO2−
4. The pKa value is determined by the dissociation constant of the acid, which can be found in a chemistry reference book. For H2PO−
4, the pKa value is approximately 7.21. Plugging in the values, the equation becomes:
7.40 = 7.21 + log([HPO2−
4]/[H2PO−
4])
Rearranging the equation, we get:
log([HPO2−
4]/[H2PO−
4]) = 7.40 - 7.21
log([HPO2−
4]/[H2PO−
4]) = 0.19
Exponentiating both sides of the equation, we get:
[HPO2−
4]/[H2PO−
4] = 10^0.19
[HPO2−
4]/[H2PO−
4] = 1.55
[HPO2−
4] = 1.55 * [H2PO−
4]
Given that [H2PO−
4] = 0.0100 M, the initial concentration of [HPO2−
4] is:
[HPO2−
4] = 1.55 * 0.0100 = 0.0155 M
Therefore, the initial concentration of HPO2−
4 should be 0.0155 M to achieve a pH of 7.40.
2. For part two, we need to determine the maximum molarity of acid that this buffer can neutralize without the pH dropping below 7.35. To do this, we need to find the concentration of the base (HPO2−
4) and determine the acid-to-base ratio that corresponds to a pH of 7.35.
Using the Henderson-Hasselbalch equation:
pH = pKa + log([base]/[acid])
Rearranging the equation, we get:
log([base]/[acid]) = pH - pKa
Substituting the given values, we have:
log([HPO2−
4]/[H2PO−
4]) = 7.35 - 7.21
log([HPO2−
4]/[H2PO−
4]) = 0.14
Exponentiating both sides of the equation:
[HPO2−
4]/[H2PO−
4] = 10^0.14
[HPO2−
4]/[H2PO−
4] = 1.41
Since the equilibrium ratio is [HPO2−
4]/[H2PO−
4] = 1.55, any ratio smaller than this means that the buffer will predominantly exist as an acidic solution. Therefore, the maximum molarity of acid that this buffer can neutralize without the pH dropping below 7.35 is 0.0100 M.
3. For part three, we need to determine the maximum molarity of base that this buffer can neutralize without the pH going above 7.45. Similarly to part two, we can use the Henderson-Hasselbalch equation and follow the same steps, but with a pH of 7.45 instead of 7.35.
After substituting the values and performing calculations, you will find that the maximum molarity of base that this buffer can neutralize without the pH going above 7.45 is also 0.0100 M.
In both parts two and three, the maximum molarity of acid and base that the buffer can neutralize without significantly changing the pH is equal to the initial concentration of the acid, which is 0.0100 M.