As a technician in a large pharmaceutical research firm, you need to produce 400. mL of 1.00 M potassium phosphate solution of pH = 7.18. The pKa of H2PO4- is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M K2HPO4 will you need to make this solution?

Use the Henderson-Hasselbalch equation.

pH = pKa + log[(base)/(acid)]
7.18 = 7.21 + log (B/A)
B/A = 0.933
and the equation B+A = 1
Solve the two equation simultaneously for B and A. Then
B*400 = mL base
A*400 = mL acid
I get approximately, but you need to do it more accurately, 207 mL acid and 193 mL base.

Well, as a technician, you're like a superhero, but instead of a cape, you wear a lab coat! Now, let's dive into the magical world of potassium phosphate solutions!

To make 400 mL of 1.00 M potassium phosphate solution of pH = 7.18, we need to find the ratio of KH2PO4 (acidic) and K2HPO4 (basic) that will give us the desired pH.

The pKa of H2PO4- is 7.21, which means that in a solution with equal amounts of KH2PO4 and K2HPO4, the pH would be close to 7.21. However, we want a pH of 7.18, so slightly more acidic than equal amounts.

First, convert the volumes of the KH2PO4 and K2HPO4 solutions to liters:
KH2PO4: 2.00 L
K2HPO4: 1.50 L

Now, we need to determine the ratio of KH2PO4 to K2HPO4. Since we want a slightly more acidic solution, let's assume we need a few more moles of KH2PO4 than K2HPO4. Let's aim for a 1.05:1 (KH2PO4:K2HPO4) ratio.

The molecular weight of KH2PO4 is 136.09 g/mol, and the molecular weight of K2HPO4 is 174.18 g/mol. So, to calculate the moles of KH2PO4 needed, we use:

(1.05 * 400 mL * 1 mol/L) / 1000 = X mol

Now, we just have to solve for X:
X = (1.05 * 400 mL * 1 mol/L) / 1000

Multiplying and dividing, we find that X = 0.42 moles.

Since the concentration of the KH2PO4 solution is 1.00 M, we can calculate the volume of KH2PO4 solution needed:

Volume of KH2PO4 solution = moles of KH2PO4 needed / concentration of KH2PO4

Volume of KH2PO4 solution = 0.42 moles / 1.00 M = 0.42 L = 420 mL

So, you will need approximately 420 mL of the 1.00 M K2HPO4 solution to make your desired potassium phosphate solution. Remember to handle it with care, because those chemicals can be a bit moody sometimes!

Now, go forth and create your magical potassium phosphate solution! Good luck, my lab coat-clad hero!

To determine the amount of 1.00 M K2HPO4 needed to make the 400 mL of 1.00 M phosphate solution with a pH of 7.18, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

In this equation, [A-] represents the concentration of the conjugate base (HPO42-) and [HA] represents the concentration of the acid (H2PO4-).

Given that the pKa of H2PO4- is 7.21 and we want a pH of 7.18, we can rearrange the equation as:

[HPO42-]/[H2PO4-] = antilog(pH - pKa)

Now, let's solve the equation step-by-step:

Step 1: Calculate the ratio of [HPO42-] to [H2PO4-].

[HPO42-]/[H2PO4-] = antilog(7.18 - 7.21)

Since the pK value is on a logarithmic scale, we need to take the anti-log (10^(7.18-7.21)).

[HPO42-]/[H2PO4-] = antilog(-0.03)

Using a scientific calculator, the antilog of -0.03 is approximately 0.50118.

[HPO42-]/[H2PO4-] = 0.50118

Step 2: Calculate the moles of KH2PO4 and K2HPO4.

Given that we want to make 400 mL of 1.00 M K2HPO4 solution, the moles of K2HPO4 would be:

Moles of K2HPO4 = 1.00 M × 0.400 L = 0.400 moles

Similarly, for KH2PO4, we'll consider it in excess and assume that all of it will react to form K2HPO4.

Moles of KH2PO4 = 1.00 M × 2.00 L = 2.00 moles

Step 3: Calculate the volume of 1.00 M K2HPO4 needed.

From step 1, we know that the ratio of [HPO42-] to [H2PO4-] is approximately 0.50118.

Therefore, the required moles of K2HPO4 can be calculated as:

0.50118 × 0.400 moles = 0.200 moles

Now, let's calculate the volume of 1.00 M K2HPO4 needed using the molarity equation:

Volume of 1.00 M K2HPO4 = Moles of K2HPO4 / Molarity

Volume of 1.00 M K2HPO4 = 0.200 moles / 1.00 M

Volume of 1.00 M K2HPO4 = 0.200 L = 200 mL

Therefore, you would need 200 mL of the 1.00 M K2HPO4 solution to make the desired 400 mL of 1.00 M potassium phosphate solution with a pH of 7.18.

To determine the amount of 1.00 M K2HPO4 solution required to make a 400 mL 1.00 M potassium phosphate solution with a pH of 7.18, we need to calculate the moles of the phosphate species present in the solution.

The given pKa value of H2PO4- (7.21) tells us that the concentration of H2PO4- is equal to the concentration of HPO4^2- at a pH of 7.21. Therefore, we can consider the potassium phosphate solution to have equal concentrations of KH2PO4 and K2HPO4.

Let's calculate the moles of H2PO4- required:

Moles of H2PO4- = Molarity * Volume
= (1.00 M) * (0.400 L)
= 0.400 moles

Since the concentration of KH2PO4 and K2HPO4 is the same, we can double the moles of H2PO4- to get the total moles of phosphate species required:

Total moles of phosphate species = 0.400 moles * 2
= 0.800 moles

Now, we need to determine the moles of K2HPO4 required to provide the desired amount of phosphate. Since the ratio of moles of phosphate to moles of K2HPO4 is 1:1, we can simply use 0.800 moles of K2HPO4.

Finally, let's calculate the volume of 1.00 M K2HPO4 solution required:

Volume of K2HPO4 = Moles of K2HPO4 / Molarity
= 0.800 moles / 1.00 M
= 0.800 L

Hence, you will need to use 0.800 liters (or 800 mL) of the 1.00 M K2HPO4 solution to make the desired 400 mL potassium phosphate solution.