CHEMISTRY 1128, EXAM I March 23, 2018 NAME 14.A phosphate buffer (H2POi HPO. ) h

Question

CHEMISTRY 1128, EXAM I March 23, 2018 NAME 14.A phosphate buffer (H2POi HPO. ) has a pH of 8.3. Which of the following will cause the a. Dissolving a small amount of Na2HPO b. Dissolving a small amount of NaH2PO c. Adding a small amount of dilute hydrochloric acid d. Adding a small amount of dilute phosphoric acid e. Making the buffer more concentrated by removing some water 15.What is [NH3] in a solution of 0.45 M NH '? K, for NHs1.8 x 10 a. 2.8 x 103M c. 1.6 x 10 M e. 2.8 x 10-10 M concentration of SO3 if the initial concentration of H2SO3 is 1.2 M? a. 7.2 x 10 M b. 2.0 x 10 M c. 1.7x 10 M d. 6.0 x 10°M e. 1.7 x 102M b. 3.5 x 10 M d. 42 x 103 M 16.Sulfurous acid is a diprotic acid, (K1 1.7 x 102, and K 6.0 10) what is the 17.The pH of Ba(OH)2 solution is 10.00. What is the H ion concentration in this solution? a. 4.0 x 1011M b. 1.0 x 1010M c. 1.6 x 101M d. 1.3x 10°M e. 10.0 M 18. Which of the following statements are true? I. A buffer can be made up by mixing "appreciable amounts of any weak acid and any weak base. Il. Consider a buffer solution prepared by combining acetic acid and sodium acetate. When a small amount of strong base is added to the buffer, the concentration of acetate ion increases. c. I and II d. None are true a.

Explanation / Answer

ANSWER:

The equilibrium reaction is given below. Let xM of NH4+ dissociate at equilibrium , we have

NH4+ <-----> NH3 + H+

0.45 - x +x +x

Kb of NH3 is given , so Ka of above reaction can be calculated as:

Kb X Ka = Kw = 10-14

Ka = 10-14 / Kb = 10-14 / 1.8 X 10-5 = 5.5 X 10-10

Now Ka = [NH3] {H+] / [NH4+] = x2 / 0.45-x

5.5 X 10-10 = x2 / 0.45-x

5.5 X 10-10 ( 0.45-x) = x2

2.47 X 10-10 - 5.5 X 10-10 x = x2

x2 + 5.5 X 10-10 x - 2.47 X 10-10  = 0

This is a quadraitic equatio its solution gives x

x = 1.6 X 10-5M

Hence option (C) IS THE CORRECT ANSWER>

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