1. A buffer was prepared by dissolving 11.7 g of KH2PO4 in 400 mL of water, and with the addition of concentrated solution of KOH the pH was adjusted to 7.10. The volume was then adjusted to 500 mL. What are the [PO43-], [HPO42-], [H2PO41-], and [H3PO4]? Use pKa values of 2.12, 7.21, and 12.7 for phosphoric acid. Show details of your calculation.
2. Which of the naturally occurring amino acids side chains are charged (>10%) at pH 2? pH 7? pH 12?
KH2PO4 + KOH ==> K2HPO4 + H2O
mols KH2PO4 = 11.7/136.08 = about 0.086
Therefore, base + acid = 0.086 (but you need to do it more accurately).
7.10 = 7.21 + log (base)/(acid)
B/A = about 0.8
B+A = about 0.086
Solve the two equations simultaneously for B and A. That will give you HPO4^2- and H2PO4^-
Use k1 to solve for H3PO4 and use k3 to solve for PO4^3-.
Post your work if you get stuck.
I don't know enough about biochem to help with #2.
I got to that part where to find the moles of KH2PO4 but why base/ acid would equal the same. this is how i went about it then i got stuck
11.7 g of KH2PO4/136.08g/mol=.085 mol of KH2PO4
then to find M:
500 ml/1000ml=.5L
.085/.5=17 M KH2PO4
then o lost myself
You don't make sense. base/acid would equal the same as what? base/acid you calculate from pH = pK2 + log (base/acid) and
B/A = approx 0.8 and I did that for you which makes base = about 0.8*A
You slipped some decimal points. 0.085/0.5L = 0.17 BUT you make sure that 0.085 is right first.
The next step you must use is to solve the two equation simultaneously.
A/B = 0.776 mols
A+B = 0.085 mols
Solve those two equation to find A and B. Then you can change to molarity if yu wish but it isn't necessary.
To answer the first question, we need to calculate the concentrations of different phosphate species in the buffer solution. Let's break it down step by step:
1. Calculate the concentration of KH2PO4:
- Molar mass of KH2PO4 = 136.09 g/mol
- Moles of KH2PO4 = mass / molar mass = 11.7 g / 136.09 g/mol = 0.0859 mol
- Concentration of KH2PO4 = moles / volume = 0.0859 mol / 0.4 L = 0.2145 M
2. Calculate the concentration of KOH added:
- As the pH was adjusted to 7.10, the KOH was added to achieve the desired pH. Since we are not given the volume or concentration of KOH added, we cannot calculate the exact concentration at this point.
3. Calculate the final volume of the buffer solution:
- The volume was adjusted to 500 mL (0.5 L) after adjusting the pH.
4. Calculate the concentration of the buffer components:
- Since the volume doubles from 0.4 L to 0.5 L, the concentration of KH2PO4 also doubles to 0.2145 M * 2 = 0.429 M.
- The buffer consists of three species: H3PO4, H2PO4-, and HPO4^2-. Let's denote their concentrations as [H3PO4], [H2PO4-], and [HPO4^2-].
Now, here comes the critical step:
5. Use the Henderson-Hasselbalch equation to calculate the concentrations of the buffer components:
- Henderson-Hasselbalch equation: pH = pKa + log([A-] / [HA])
- In this case, we can assume that H2PO4- acts as the acid and HPO4^2- acts as the conjugate base. Therefore, we will use pKa2 = 7.21.
For H2PO4- (acid):
- pH = 7.10 (given)
- pKa = 7.21 (given)
- [A-] = [HPO4^2-]
- [HA] = [H2PO4-]
- 7.10 = 7.21 + log([HPO4^2-] / [H2PO4-])
Similarly, for HPO4^2- (conjugate base):
- pH = 7.10 (given)
- pKa = 7.21 (given)
- [A-] = [HPO4^2-]
- [HA] = [H2PO4-]
- 7.10 = 7.21 + log([HPO4^2-] / [H2PO4-])
Now solve for [HPO4^2-] and [H2PO4-].
6. Calculate [PO43-], [H3PO4], and [H2PO4-]:
- [HPO4^2-] = 10^-(7.21 - 7.10 + log([HPO4^2-] / [H2PO4-]))
- [PO43-] = [HPO4^2-]
- [H3PO4] = [H2PO4-] = [KH2PO4] - [HPO4^2-]
Using these calculations, we can determine the concentrations of [PO43-], [HPO4^2-], [H2PO4-], and [H3PO4] in the buffer solution.
For the second question, we need to consider the pKa values and the charge of side chains of different amino acids at different pH levels. Here are the groups of amino acids side chains based on their charge:
1. Charged (>10%) at pH 2:
- Aspartate (Asp or D)
- Glutamate (Glu or E)
2. Charged (>10%) at pH 7:
- Lysine (Lys or K)
- Arginine (Arg or R)
- Histidine (His or H)
3. Charged (>10%) at pH 12:
- Aspartate (Asp or D)
- Glutamate (Glu or E)
- Lysine (Lys or K)
- Arginine (Arg or R)
Therefore, at pH 2, the amino acids with side chains Aspartate (Asp or D) and Glutamate (Glu or E) will be charged (>10%). At pH 7, the amino acids with side chains Lysine (Lys or K), Arginine (Arg or R), and Histidine (His or H) will be charged (>10%). At pH 12, the same amino acids as in pH 7 (Lysine, Arginine, and Histidine) will still be charged (>10%), along with Aspartate (Asp or D) and Glutamate (Glu or E).