You have at your lab bench the following chemicals: NaH2PO4(s), Na3PO4(s), NH4Cl(s), CH3COONa (s), 1.00 M NaOH, 1.00 M HCl, and deionized water. You also have standard glassware available. Describe how you would make 1.0 L of the buffer with a pH of 7.50 using only the materials listed above and deionized water so that the concentration of the acid component in the buffer is 0.100 M. Give the amounts of the pure materials in grams if they are solid or mL if they are liquid.

Acid pKa
H3PO4 2.15
H2PO4- 7.20
HPO42- 12.3
NH4+ 9.24
CH3COOH 4.75
Correct Answer: 36 g NaH2PO4, 0.20 L NaOH, water to get 1.00 L total.

I know that I have to use H2PO4- because it has the closest pH to our desired pH. In the answer key, my prof gave us I don't understand why we add NaOH and how do I find the amount of NaOH????? Please HELP!!

What must base be for acid to be 0.1M?

7.5 = 7.2 + log b/a
7.5= 7.2 + log (b/0.1)
base = 0.2M

Then H2PO4^- + OH^- ==> HPO4^2- + H2O
I...........?...............0.................0..........
add.......................x..............................
C.........-x.............-x....................x
E.........0.1............0.....................x

But you know x (the base) must be 0.2M so you work backwards on the ICE chart and C is -0.2M; therefore, I (initial) must be 0.1 + 0.2 = 0.3 M. So you weigh out 0.3 mol NaH2PO4 and that in 1.0 L is 0.3M. You add 0.2 mol NaOH (it is 1.0 M so that is 0.2L) and tht should give you a pH of 7.5 BUT you always check it.
..................H2PO4^- + OH^- ==> HPO4^2- + H2O
I..................0.3................0................0
add...............................0.2....................
C................-0.2..........-0.2....................0.2
E..................0.1..............0...................0.2

pH = 7.2 + log (b/a)
pH = 7.2 + log (0.2/0.1)
pH = 7.2 + log 2 = 7.2 + 0.3 = 7.5
Voila!

Well, making a buffer can sometimes be as confusing as trying to understand a clown's punchline. But don't worry, I'm here to help! In this case, NaOH is added to the buffer to help adjust the pH.

To find the amount of NaOH needed, we can use the Henderson-Hasselbalch equation, which is like a mathematical joke that chemists love to tell. The equation is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base (in this case, H2PO4-) and [HA] is the concentration of the acid (in this case, NaH2PO4).

Here, the pH is given as 7.50 and the pKa of H2PO4- is 7.20. Since we want a buffer with a pH of 7.50, that means the [A-] and [HA] concentrations should be equal.

Now, we have the desired concentration of the acid component, which is 0.100 M NaH2PO4. Since the NaH2PO4 is a solid, we need to calculate the amount in grams. The molar mass of NaH2PO4 is roughly 119.98 g/mol.

To get 0.100 M concentration in 1.0 L, we would need 0.100 moles of NaH2PO4. So, the mass of NaH2PO4 would be:

0.100 mol * 119.98 g/mol = 11.998 g ≈ 12 g NaH2PO4.

Now, we need to determine the amount of NaOH to add. Since we want to have equal concentrations of [A-] and [HA], we need to find the volume of NaOH to add that will react with the 0.100 M NaH2PO4.

In this case, 1 mole of NaOH (mol wt. roughly 39.99 g/mol) reacts with 1 mole of NaH2PO4. So, the volume of NaOH we need is:

0.100 mol NaH2PO4 * (39.99 mL NaOH / 1 mol NaOH) ≈ 4 mL

But this calculation gives us the amount needed for only 0.1 L of solution. Since we want to make 1.0 L of buffer, we need to scale up the amounts by a factor of 10.

So, the final amounts would be approximately 12 g NaH2PO4 and 40 mL (or 0.04 L) NaOH. Of course, don't forget to add deionized water to reach a total volume of 1.0 L.

And there you have it! You've successfully made your buffer solution. Now go out there and embrace the science of clownistry!

To prepare the buffer with a pH of 7.50 using NaH2PO4 as the acid component, you need to create a solution that contains both the acid (H2PO4-) and its conjugate base (HPO42-). The Henderson-Hasselbalch equation can be used to calculate the ratio of the two components required to achieve the desired pH.

Henderson-Hasselbalch Equation: pH = pKa + log ([HPO42-] / [H2PO4-])

In this case, the desired pH is 7.50 and the pKa of the acid component H2PO4- is 7.20.

pH = 7.50
pKa = 7.20

To solve for the ratio [HPO42-] / [H2PO4-], we need to rearrange the Henderson-Hasselbalch equation:

[HPO42-] / [H2PO4-] = 10^(pH - pKa)

[HPO42-] / [H2PO4-] = 10^(7.50 - 7.20)
[HPO42-] / [H2PO4-] = 10^0.30
[HPO42-] / [H2PO4-] = 1.995

Since we want the concentration of the acid component (H2PO4-) to be 0.100 M, we can assign a value of 0.100 to [H2PO4-]. Using the ratio from above, we can then solve for [HPO42-]:

[HPO42-] = [H2PO4-] * 1.995 = 0.100 * 1.995 = 0.1995 M

Now, we can calculate the number of moles of H2PO4- required to prepare 1.0 L of the buffer with a concentration of 0.100 M:

moles of H2PO4- = volume (L) * concentration (M) = 1.0 * 0.100 = 0.100 mol

Since the molar mass of NaH2PO4 is 119.98 g/mol, we can calculate the mass of NaH2PO4 required:

mass of NaH2PO4 = moles * molar mass = 0.100 * 119.98 = 11.998 g

Therefore, you would need to add 11.998 grams of NaH2PO4 to the lab bench to prepare the buffer.

Now, we need to determine how much NaOH to add. NaOH will be used to adjust the pH of the buffer solution. Since NaOH is a strong base, it will react with H2PO4- to form additional HPO42- and water:

H2PO4- + OH- -> HPO42- + H2O

To prepare a 1.0 L buffer solution, you would add deionized water and NaOH solution to achieve a final volume of 1.0 L. The exact amount of NaOH required will depend on the initial pH of the buffer solution, which is determined by the concentration of the acid component H2PO4-.

In the provided answer, the instructor has the correct ratio of NaH2PO4 to NaOH to achieve a pH of 7.50. The amount of NaOH required, in this case, is given as 0.20 L. However, the exact calculation to determine this volume is not provided. It appears that in this example, the instructor has used an experimental approach or a pre-determined value to determine the amount of NaOH required.

To accurately determine the exact volume of NaOH required, you would need to perform extensive titration experiments or use a buffer calculator that takes into account the pKa, desired pH, and initial concentrations.

In summary, to make 1.0 L of the buffer with a pH of 7.50 and an acid component concentration of 0.100 M, you would need to add 36 grams of NaH2PO4 to the lab bench along with an appropriate volume of NaOH solution. The exact volume of NaOH required would need to be determined experimentally or with the help of a buffer calculator.

To create a buffer solution with a pH of 7.50 using H2PO4- as the acid component, you need to utilize the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid])

In this case, the pKa of H2PO4- is 7.20. We want the concentration of the acid component (H2PO4-) to be 0.100 M. Therefore, we can rearrange the equation to solve for [base]:

[base] = [acid] * 10^(pH - pKa)

Substituting the values, we get:

[base] = 0.100 M * 10^(7.50 - 7.20)

Simplifying gives:

[base] ≈ 0.158 M

Now, we need to determine how much NaH2PO4 to add to achieve this concentration. NaH2PO4 dissociates into H2PO4- and Na+ in solution, so the concentration of H2PO4- will be equal to the concentration of NaH2PO4.

Let's assume we have 'x' moles of NaH2PO4, which will give us the desired concentration of H2PO4- in moles per liter (M). Since 1 mole of NaH2PO4 corresponds to 1 mole of H2PO4-, the concentration of H2PO4- will also be 'x' M.

Therefore, we have:

0.158 M = x M

Solving for 'x', we find:

x ≈ 0.158 M

To convert this molar concentration to grams, we need to calculate the molecular weight of NaH2PO4. The molecular weight of Na is 22.99 g/mol, H is 1.01 g/mol, and PO4 is 94.97 g/mol. Adding these together gives a molecular weight of NaH2PO4 of approximately 119.96 g/mol.

Now, we can calculate the mass (in grams) of NaH2PO4 required:

mass = moles * molecular weight
mass = 0.158 M * 1 L * 119.96 g/mol
mass ≈ 18.94 g

Therefore, to make 1.0 L of the buffer solution at pH 7.50 with a concentration of the acid component (H2PO4-) at 0.100 M, you would need approximately 18.94 grams of NaH2PO4.

Now, let's address the addition of NaOH. Since NaOH is a strong base, it will react with the acidic component, H2PO4-, to produce water and the conjugate base (HPO42- in this case). The reaction is as follows:

H2PO4- + OH- → HPO42- + H2O

By adding NaOH, we increase the concentration of OH-, which will drive the reaction to the right, resulting in an increase in the concentration of the base (HPO42-) and a corresponding decrease in the concentration of the acid (H2PO4-). This helps to maintain the desired pH of the buffer solution.

The exact amount of NaOH needed depends on the initial concentration of the acid component and desired buffer capacity. In the given answer, your professor determined that adding 0.20 L of 1.00 M NaOH was sufficient to achieve the desired pH and concentration of the acid component.

In summary, to make 1.0 L of the buffer solution with a pH of 7.50 and a concentration of the acid component (H2PO4-) at 0.100 M, you would need to add approximately 18.94 grams of NaH2PO4 and 0.20 L of 1.00 M NaOH to deionized water to reach a total volume of 1.00 L.