CHEM 1403 ________________________
COOP 5 (name)
Dr. Robert E. Belford
1. (25 pts) What is the pH of a solution which is 1.0 x 10-4M in HNO3?
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2. (25 pts) What is the hydonium ion concentration of a solution which is 3 M in hydrocyanic acid (HCN), Ka (HCN) = 4.9 x 10-10?
As KA<<<[HCN]i you can ignore the extent of reaction.
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3. Consider a solution which is 3M in sodium cyanide (NaCN), Ka (HCN) = 4.9 x 10-10 .
Kb=Kw/Ka = 10-14/(4.9
x 10-10 )=2.04 x 10-5
a. (20 pts) What is the hydroxide ion concentration?
As 100Kb<[CN-]i you can ignore the extent of reaction.
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b. (5 pts) What is the pOH?
pOH- = -log[OH-] = 2.1
c. (5 pts)What is the pH?
pH =14- pOH = 11.9
4. Determine via chemical equation whether aqueous solutions of the following salts are neutral, acidic or basic.(20 pts)
a. NaCl
Neutral, Na+ is the cation from a strong base
Cl- is the anion from a strong acid
No reaction
b. NH4Cl Acidic as ammonium is the conjugate acid of the weak base ammonia and thus it is a weak acid
NH4+ + H2O --> NH3 + H3O+
5. Consider the titration of 50.00 mL of 0.100M acetic acid (CH3COOH) with 0.200 M Soidum Hydroxide (NaOH). What is the pH when the following amounts of base have been added? You may check your work with the Acic/Base virtual lab.
a. 0.00 mL
Treat as Weak Acid
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b. 10.00 mL
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c. 12.50 mL
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d. 15.00 mL
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e. 24.00 mL
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f. 25.00 mL
This is the Equivalence Point, treat as the salt of the weak acid

g. 26.00 mL, [OH-] is determined by moles excess base (1 ml in excess of that required to neutralize all of the acid)

h. 30.0 mL

6. A buffer is prepared by mixing 17.5 g of acetic acid (fw = 60.02 g/mol) with 15.8 g sodium acetate (fw = 82.02g/mol), and diluting to 500.0 mL. What is the pH of the buffer?

7. What must be the ratio of acetic acid to sodium acetate to prepare a buffer whose pH = 4.81?

8. The solubility of gold chloride (AuCl3(s) <=> Au3+ + 3Cl- ) in water is 1.04 x 10-6 mol/L. Calculate the value of the solubility-product constant, Ksp, for gold chloride.
Ksp
= X(3X)3 = 33(X)4=33(1.04 x 10-6)4
= 3.16x10-23
9. The solubility-product constant for lead iodide (PbI2 <=> Pb2+ + 2I- ) is Ksp = 7.1 x 10-9. Calculate the molar solubility of lead iodide in otherwise pure water.

10. Calculate the molar solubility of lead iodide (PbI2 <=> Pb2+ + 2I- , Ksp = 7.1 x 10-9) in a solution containing 0.10 M potassium iodide (KI), a very soluble salt.
