Pre-Lab Assignment Hydrogen Phosphate Buffer System This pre-lab assignment assi

Question

Pre-Lab Assignment Hydrogen Phosphate Buffer System This pre-lab assignment assignment is not completed before the beginning of the lab is mandatory. You will receive a zero for the experiment if the pre-lab 1. Calculat e the masses of the conjugate acid and base needed to prepare each of the six buffer solutions n the left. (pK, of HPO,6.70) Assume the availability of the phosphate salts listed in the table o listed in the table on the right. Fill out the worksheet below. Show all work on separate pages for full credit. Molar mass (g mol-1) 136.09 174.2 137.99 268.09 KH2PO (0.25 M, 100 mL) pH 6.50 (Solution I) pH 6.70 (Solution 2) PH 6.90 (Solution 3) (0.25 M, 100 mL.) pH 6.50 (Solution 4) pH 6.70 (Solution 5) pH 6.90 (Solution 6) Salt KH2PO K2HPO NaH PO H2O Na HPO,.7 H2O pH 6.70 pH 6.90 PH 6.50 Moles of H2PO D S mble Moles of HPo2 Solution 1 Solution 2 Solution 3 1.32 Grams of KH PO Grams of K2HPO Solution 4 Solution 5 Solution 6 Grams of NaH,PO, H0 Grams ofNaz HPO. . 7H20 255 3s 09 2. Suppose you added 2.00 mL of 0.250 M HCI 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. 59

Explanation / Answer

2.

From question 1:

At pH 6.70

Moles of H2PO4- = 0.013 mole
Moles of HPO42- = 0.013 mole

Now, you added 2.00 mL of 0.250 M HCl

Molarity = Moles/Liter
Moles = Molarity x Liter
           = 0.250 M x (2/1000) L
           = 0.250 M x 0.002 L
           = 0.0005 moles

Now, 0.0005 moles of HCl will react with 0.0005 moles of HPO42- to form 0.0005 moles of H2PO4-. So, moles of HPO42- will decrease by 0.0005 moles and moles of HPO42- will increase by 0.0005 moles.

So, moles after addition of HCl are

Moles of H2PO4- = 0.013 mole + 0.0005 mole = 0.0135 mole
Moles of HPO42- = 0.013 mole – 0.0005 mole = 0.0125 mole

Now.

According to Henderson-Hesselbalach equation

pH = pKa + log { [salt] / [acid] }

pH = pKa + log { [HPO42-] / [H2PO4-] }

pH = pKa + log { moles of HPO42- / moles of H2PO4- }

pH = 6.70 + log { 0.0125 / 0.0135 }

pH = 6.70 + log { 0.926 }

pH = 6.70 + (- 0.03)

pH = 6.67

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