pKa1 = 2.148, pKa2 = 7.198, pKa3 = 12.375
You wish to prepare 1.000L of a 0.0100M Phosphate buffer at pH7.55. To do this, you choose to use mix the two salt fomrs involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?
mass NaH2PO4 and Na2HPO4?
What other combination of phosphoric acid and /or its salt could be mixed to prepare this buffer?
*Check all that apply*
a) H3PO4 and NaH2PO4
b) H3PO4 and Na2HPO4
c) H3PO4 and Na3PO4
d) NaH2PO4 and Na3PO4
e) Na2HPO4 and Na3PO4
what weight of NaH2Po4 and na2hp04 would be required to prepare 500 ml of a buffer solution of pH 7.45 that has an ionic strength of 0.1
Well, preparing a buffer solution is like preparing a perfect blend of ingredients for a delicious dish. Let's calculate the mass of each salt that we need!
To achieve a pH of 7.55, we need to choose a combination of NaH2PO4 and Na2HPO4 that will be able to handle our taste preferences. Sorry, I mean pH requirements.
We know that the pKa2 value is 7.198, which is close to our desired pH of 7.55. This means we want a significant amount of both the acidic form (NaH2PO4) and the basic form (Na2HPO4) in our buffer solution.
Now, let's get down to the nitty-gritty calculations. We can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
For the first pKa (pKa1), we don't really care because it will have negligible influence on the pH in this case.
Now, let's focus our attention on the second pKa (pKa2).
pH = pKa2 + log([Na2HPO4]/[NaH2PO4])
We want to prepare a 0.0100 M buffer, so let's assign that concentration to [HA] (NaH2PO4).
0.0100 = 7.198 + log([Na2HPO4]/[0.0100])
Rearranging the equation, we get:
log([Na2HPO4]) - log([0.0100]) = 7.198 - 0.0100
log([Na2HPO4]) = 7.188
Using the properties of logarithms, we find:
[Na2HPO4] = 10^(7.188) = 1.555
Therefore, the concentration of NaH2PO4 will be 0.0100 M, and the concentration of Na2HPO4 will be 1.555 M.
To determine the mass of each salt, we can use the following formula:
mass = concentration × volume × molar mass
For NaH2PO4:
mass of NaH2PO4 = 0.0100 M × 1.000 L × (mass of 1 mol of NaH2PO4)
And for Na2HPO4:
mass of Na2HPO4 = 1.555 M × 1.000 L × (mass of 1 mol of Na2HPO4)
Now, I'm not great with numbers, but you can plug those values into the equation and solve for the masses. Don't forget to check your calculations, just like a chef tastes their food before serving!
As for the other combinations of phosphoric acid and/or its salt that could be mixed to prepare this buffer, let's go through the options:
a) H3PO4 and NaH2PO4:
This combination is suitable for preparing a buffer at a pH close to the first pKa value (2.148). Since our target pH is 7.55, this combination is not suitable.
b) H3PO4 and Na2HPO4:
This combination is suitable for preparing a buffer at a pH close to the second pKa value (7.198). Since our target pH is 7.55, this combination is also not suitable.
c) H3PO4 and Na3PO4:
This combination is suitable for preparing a buffer at a pH close to the third pKa value (12.375). Since our target pH is 7.55, this combination is not suitable.
d) NaH2PO4 and Na3PO4:
NaH2PO4 is suitable for preparing buffers close to the first pKa, while Na3PO4 is suitable for preparing buffers close to the third pKa. Since our target pH is close to the second pKa, this combination is not suitable.
e) Na2HPO4 and Na3PO4:
Na2HPO4 is the basic form suitable for preparing buffers close to the second pKa, while Na3PO4 is suitable for preparing buffers close to the third pKa. Since our target pH is in the range of the second pKa, this combination is a potential option for preparing the buffer.
So, the correct answer from the options given would be:
e) Na2HPO4 and Na3PO4
Please keep in mind that preparing buffer solutions is like chemistry cooking, and precise calculations are essential for getting the desired results.
To calculate the mass of each salt (NaH2PO4 and Na2HPO4) needed to prepare the phosphate buffer, we first need to understand the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([salt]/[acid])
In this case, our desired pH is 7.55. We can find the pKa value closest to this pH, which is pKa2 = 7.198. This corresponds to the second ionization step of the phosphate buffer system: H2PO4- (acid) ⇌ HPO4^2- (salt).
Using the Henderson-Hasselbalch equation, we can rewrite it as:
7.55 = 7.198 + log ([HPO4^2-]/[H2PO4-])
Now we can solve for the ratio [HPO4^2-]/[H2PO4-].
Taking the anti-logarithm of both sides, we have:
[HPO4^2-]/[H2PO4-] = antilog(7.55 - 7.198)
Using the antilog function on a calculator, we find:
[HPO4^2-]/[H2PO4-] ≈ 4.473
Since the ratio of the salt to the acid is approximately 4.473, we can set up the following equation to calculate the concentration of each species in terms of molarity and volume:
For NaH2PO4:
0.0100 M = [NaH2PO4] / (1.000 L)
For Na2HPO4:
0.0100 M = [Na2HPO4] / (1.000 L)
Now we can solve for the mass of each salt. The molar mass of NaH2PO4 is approximately 119.98 g/mol, while the molar mass of Na2HPO4 is approximately 141.96 g/mol.
For NaH2PO4:
mass NaH2PO4 = 0.0100 M * 1.000 L * molar mass of NaH2PO4
For Na2HPO4:
mass Na2HPO4 = 4.473 * mass NaH2PO4
Calculating the masses, we have:
mass NaH2PO4 ≈ 0.0100 g/mol * 1.000 L * 119.98 g/mol = 1.1998 g
mass Na2HPO4 ≈ 4.473 * 1.1998 g ≈ 5.359 g
Therefore, you will add approximately 1.1998 g of NaH2PO4 and 5.359 g of Na2HPO4 to prepare the phosphate buffer.
Now let's consider the other combinations of phosphoric acid and/or its salt that could be mixed to prepare this buffer. We need to look for combinations that will provide the desired pH when the acid and salt are mixed in appropriate ratios.
Based on the given pKa values, phosphoric acid (H3PO4) has pKa1 = 2.148, pKa2 = 7.198, and pKa3 = 12.375.
By examining the Henderson-Hasselbalch equation, we can determine the possible combinations by considering the acid-salt pairings and comparing their pKa values with the desired pH of 7.55.
a) H3PO4 and NaH2PO4:
The pKa1 value of H3PO4 is lower than the desired pH of 7.55. This combination will not provide the desired pH, so it is not a possible option.
b) H3PO4 and Na2HPO4:
The pKa2 value of H3PO4 is close to the desired pH of 7.55. This combination can potentially provide the desired pH and is a possible option.
c) H3PO4 and Na3PO4:
The pKa3 value of H3PO4 is higher than the desired pH of 7.55. This combination will not provide the desired pH, so it is not a possible option.
d) NaH2PO4 and Na3PO4:
Neither NaH2PO4 nor Na3PO4 are acids, so this combination will not be able to provide the desired pH. Therefore, it is not a possible option.
e) Na2HPO4 and Na3PO4:
Both Na2HPO4 and Na3PO4 are salts, and they do not involve any acidic species. This combination will not provide the desired pH value, so it is not a possible option.
In conclusion, the other combination of phosphoric acid and/or its salt that could be mixed to prepare this buffer is b) H3PO4 and Na2HPO4.
For second Part :
B,C,D
You have two unknowns so you need two equations.
pH = pK2 + log(base)/(acid)
Substitute the number and solve for base/acid for one equation.
The second is base + acid = 0.1M
Post your work if you get stuck.