1. Calculate the masses of the conjugate acid and base needed to prepare each of

Question

1. Calculate the masses of the conjugate acid and base needed to prepare each of the six buffer solutions listed in the table on the left. (pKa of H2PO 6.70) Assume the availability of the phosphate salts listed in the table on the right. Fill out the worksheet below. Show all work on separate pages for full credit. Salt KH2PO K2HPO4 NaH2P H20 Na2HPO,.7H20 Molar mass (gmol-1) 136.09 174.2 137.99 268.09 KH2PO4/K2HPO (0.25 M, 100 mL) (0.25 M, 100 mL) pH 6.50 (Solutio pH 6.50 (Solution 4) pH 6.70 (Solution 2) pH 6.70 (Solution 5) pH 6.90 (Solution 3) pH 6.90 (Solution 6) H 6.50 H 6.70 H 6.90 Moles of H2PO Moles of HPO2 Solution Solution 2 Solution 3 Grams of KH2PO Grams of K2HPO Solution 4 Solution 5 Solution 6 Grams of NaH2P H20 Grams of Na2HPO4-7 H20 2. Suppose you added 2.00 mL of 0.250 M HCl to the pH 6.70 hydrogen phosphate buffer that was made in Question 1. What is the pH of the resulting buffer solution? Show all work.

Explanation / Answer

I will tell you how t calculate the first solution and then try to do the other 5 yourself using the same procedure I put here but changing the values of pH only.

You want to prepare 100 mL of a 0.25 M buffer of H2PO4-/HPO42-, to do this, first calculate the ratio of this buffer:

pH = pKa + log(H2PO4-/HPO42-)
6.5 = 6.70 + log(H2PO4-/HPO42-)
-0.2 = log(H2PO4-/HPO42-)
10-0.2 = H2PO4-/HPO42-
H2PO4-/HPO42- = 0.63
H2PO4- = 0.63HPO42-

Now, we know that H2PO4- + HPO42- = Buffer = 0.25 M so:
0.25 = 0.63HPO42- + HPO42-
0.25 = 1.63HPO42-
HPO42- = 0.1533 M
H2PO4- = 0.63*0.1533 = 0.0966 M

We have the concentrations, now let's calculate the moles, and finally the mass:
moles H2PO4- = 0.0966 mol/L * 0.1 L = 0.00966 moles
moles HPO42- = 0.1533 mol/L * 0.1 L = 0.01533 moles

Now the mass:
mass KH2PO4 = 0.00966 * 136.09 = 1.3146 g
mass K2HPO4 = 0.01533 * 174.20 = 2.6705 g

This is for solution 1. For solution 4, the mass would be:
mass NaH2PO4 = 0.00966 * 137.99 = 1.3329 g
mass Na2HPO4 = 0.01533 * 268.09 = 4.1098 g

Use the same procedure for solution 2, 3, 5 and 6 and you will get the final results.

For question 2, all you have to do is use the HH equation with the moles you obtain for that solution, and then:
pH = pKa + log (moles H2PO4- - moles OH- / moles HPO4 + moles OH-)

This will give you the pH

Hope this helps

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