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?
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.
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.
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)
Neutral, Na+ is the cation from a strong base
Cl- is the anion from a strong acid
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
b. 10.00 mL
c. 12.50 mL
d. 15.00 mL
e. 24.00 mL
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.