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 = 7.06. The pKa of H_2PO_4}^- is 7.21. You have 2.00 L of 1.00 M KH_2PO_4 solution and 1.50 L of 1.00 M K_2HPO_4 solution, as well as a carboy of pure distilled H_2O.

How much 1.00 M KH_2PO_4 will you need to make this solution?

Use the Henderson-Hasselbalch equation and solve for (base)/(acid) ratio (this is equation 1).

The problem gives you equation 2 which is
(acid) + (base) = 1M

Solve these two equations simultaneously to determine mols acid and mols base needed; then translate this into volume of the two materials at hand. Post your work if you get stuck. I zipped through this very quickly and it appears you will need about 117 mL acid and about 83 mL of the base but I could have made a math error easily. You should confirm those numbers. .

To calculate how much 1.00 M KH2PO4 is needed to make the 200 mL solution, we need to consider the pH and the pKa of the H2PO4– ion.

The formula for a phosphate buffer solution is given by:

pH = pKa + log([H2PO4–]/[HPO4^2–])

Given that pH = 7.06 and pKa = 7.21, we can rearrange the equation to solve for the ratio [H2PO4–]/[HPO4^2–]:

[H2PO4–]/[HPO4^2–] = 10^(pH – pKa)

[H2PO4–]/[HPO4^2–] = 10^(7.06 – 7.21)

[H2PO4–]/[HPO4^2–] = 0.630957

Now, let's assume x is the volume (in liters) of 1.00 M KH2PO4 needed to make the buffer solution. Then, (2.00 - x) L of 1.00 M KH2PO4 will be left. Since the molarities of KH2PO4 and K2HPO4 are the same, the resulting H2PO4–/HPO42– ratio should be equal to the ratio of the remaining KH2PO4 volume to the K2HPO4 volume. So, we have:

[H2PO4–]/[HPO4^2–] = (2.00 - x) L / 1.50 L

0.630957 = (2.00 - x) / 1.50

Solving for x:

x = 2.00 - (0.630957 * 1.50)

x ≈ 1.055 L

Therefore, you would need approximately 1.055 L of the 1.00 M KH2PO4 solution to make the 200 mL phosphate buffer solution with a pH of 7.06.