As a technician in a large pharmaceutical research firm, you need to produce 200.mL of 1.00 M a phosphate buffer solution of pH = 7.19. 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 KH2PO4 will you need to make this solution?

Question

As a technician in a large pharmaceutical research firm, you need to produce 200.mL of 1.00 M a phosphate buffer solution of pH = 7.19. 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 KH2PO4 will you need to make this solution?

Answer 1

You need 200 mL x 1M so

base(b) + acid(a) = 0.2 mols.
That's equation 1.

7.19= 7.21 + log b/a
b/a = ? and that's equation 2

Solve those two equations to find a and b then substitute each into M = mol/L to find the individual amounts for a and b.
Post your work if you get stuck.
I went through this in a hurry and came up with about 98 mL base and 103 mL acid but these are not exact.

Answer 2

To determine the amount of 1.00 M KH2PO4 solution you need to make a 200 mL of 1.00 M phosphate buffer solution with a pH of 7.19, we need to use the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is as follows:

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

In this case, the acid HA is H2PO4- and the conjugate base A- is HPO42-.

Given that pKa = 7.21, and pH = 7.19, we can rearrange the equation to solve for the ratio [A-]/[HA]:

[A-]/[HA] = 10^(pH - pKa)

[A-]/[HA] = 10^(7.19 - 7.21)

[A-]/[HA] = 10^(-0.02)

[A-]/[HA] = 0.6309573445

Now, let's assume that x is the amount of 1.00 M KH2PO4 solution (in mL) that we need to make the buffer solution.

So, to make the buffer solution, the amount of the conjugate base A- (HPO42-) in the KH2PO4 solution will be (0.6309573445 * x) mL.

And the amount of the acid HA (H2PO4-) in the KH2PO4 solution will be ((1 - 0.6309573445) * x) mL.

To make a 200 mL buffer solution, the total volume of the KH2PO4 solution and water is 200 mL, so:

(0.6309573445 * x) + ((1 - 0.6309573445) * x) + volume of water = 200 mL

Simplifying the equation gives:

x = (200 mL * (1 - 0.6309573445)) / (0.6309573445 + (1 - 0.6309573445))

x = (200 mL * 0.3690426555) / 1

x = 73.81 mL

Therefore, you will need 73.81 mL of the 1.00 M KH2PO4 solution to make the 200 mL of 1.00 M phosphate buffer solution.