Quiz 4 (08-09)
Chem 171
20 points
Name_____________________
1) Water is a polar molecule because:
a) it is a bent molecule and has ionic bonds.
c) it is composed of 3 atoms.
b) it has a small molecular weight.
d) it is a bent molecule and has polar covalent bonds.
2) Predict the products of the following reactions and provide a balanced molecular equation. Be sure to include the
phase. If there is no net reaction, indicate as such.
a) KNO3 (aq) + CuCl2 (aq)  __no net reaction OR 2 KCl (aq) + Cu(NO3)2 (aq) ____
b) Li2SO4 (aq) + BaCl2 (aq)  __BaSO4 (s) + 2 LiCl (aq)___
c) HBr (aq) + KOH (aq)  ___KBr (aq) + H2O (l)__
3) What is the molarity of Na2HPO4 in a solution if 0.35 grams of Na2HPO4 is in a final volume of 500 mL? ___4.9 _mM_
+
What is the molarity of Na ?
-3
0.35 g x mol/141.9 g = 2.47 x 10 mol Na2HPO4 in 500 mL or 0.500 L
-3
___9.87 mM___
-3
2.47 x 10 mol/0.500 L = 4.9 x 10 M or 4.9 mM
+
-3
+
Each mole of Na2HPO4 yields 2 mol Na so there are 4.94 x 10 mol Na
-3
-3
4.94 x 10 mol/ 0.500 L = 9.87 x 10 M or 9.87 mM
4) A solution of lead(II)nitrate is added to a solution of potassium sulfide. A lead(II) sulfide precipitate forms.
Provide the molecular equation, the total ionic equation and the net ionic equation for this reaction
Molecular: Pb(NO3)2 (aq) + K2S (aq) → 2 KNO3 (aq) + PbS (s)
2+
+
-2
+
Total ionic: Pb (aq) + 2 NO3 (aq) + 2 K (aq) + S (aq) → 2 NO3 (aq) + 2 K (aq) + PbS (s)
2+
-2
Net ionic: Pb (aq) + S (aq) → PbS (s)
5) A solution of nitrous acid (HNO2) is added to a solution of sodium hydroxide.
Provide the molecular equation, the total ionic equation and the net ionic equation for any reaction that occurs.
(Notice, this is a weak acid)
HNO2 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)
+
+
HNO2 (aq) + Na (aq) + OH (aq) → Na (aq) + NO2 (aq) + H2O (l)
HNO2 (aq) + OH (aq) → NO2 (aq) + H2O (l)
-5
6) What is the pH of a solution that has a proton concentration of 5.34 x 10 M?
+
-5
pH = -log[H ] = -log (5.34 x 10 )
pH is 4.27
7) What is the proton concentration of a solution with a pH of 9.4?
+
-pH
[H ] = 10
-9.4
= 10
= 3.98 x 10
-10
8) Consider the reaction between lead (II) nitrate and potassium sulfide (#4 above)
a) How much lead (II) sulfide would form if 150 mL 0.0500 M lead (II) nitrate was added to
350 mL 0.030 M potassium sulfide?
_____1.79g_____
b) What is the concentration of nitrate ion after reaction?
______0.03 M______
c) What is the concentration of sulfide ion after reaction?
______6.0 x 10 M___
-3
Determine limiting reagent:
-3
-3
0.150 L x 0.0500 mol Pb(NO3)2/L = 7.5 x 10 mol x 1mol PbS/1 mol Pb(NO 3)2 = 7.5 x 10 mol PbS
-2
-2
0.350 L x 0.0300 mol K2S/L = 1.05 mol mol x 1 mol PbS/1 mol K2S = 1.05 x 10 mol PbS
-3
Shows that Pb(NO3)2 is limiting and 7.5 x 10 mol PbS would form
-3
This is 7.5 x 10 mol x 239.2 g/mol = 1.79 g
Amount of nitrate after reaction:
This did not precipitate so calculate new concentration
-3
-2
0.150 L x 0.0500 mol Pb(NO3)2/L = 7.5 x 10 mol Pb(NO3)2 x 2 mol NO3/mol Pb(NO3)2 = 1.5 x 10 mol NO3
-2
-2
This is present in 0.500 L
1.5 x 10 mol/0.500 L = 3.0 x 10 M
Amount of sulfide after reaction:
Some of this precipitated. How much?
Determine how much sulfide reacted with lead
-3
0.150 L x 0.0500 mol Pb(NO3)2/L = 7.5 mol mol Pb(NO3)2 x (mol K2S/1 mol Pb(NO3)2) x 1 mol S/1 mol K2S
-3
+2
= 7.5 x 10 mol Pb reacted
Determine how much sulfide was there initially
-2
-2
0.350 L x 0.0300 mol K2S/L = 1.05 mol mol x 1 mol S/1 mol K2S = 1.05 x 10 mol S
-2
The amount of sulfide after precipitation is (initial – precipitated) = 1.05 x 10
-3
-3
in 0. 500 L
3.0 x 10 /0.50 = 6.0 x 10 M
-3
- 7.5 x 10
-3
= 3.0 10 mol
9) If 0.200 L of 0.20 M Ca(OH)2 is mixed with 0.300 L of 0.25 M HCl, will the resulting solution be acidic or basic?
Ca(OH)2 (aq) + 2 HCl (aq) → CaCl2 (aq) + H2O (l)
-2
0.200 L base x 0.2 mol/L x 2 mol H2O/1 mol Ca(OH)2 = 8 x 10 H2O
0.300 L acid x 0.25 mol/L x 2 mol H2O/2 mol HCl = 7.5 x 10
-2
H2O
Base is in excess so the solution would be basic.
10) Provide the oxidation number of the designated atom.
-
a) Cl in ClO2 _+3__
b) N in N2 _0___
d) C in CO2H2 _+2___
e) Ca in Ca
2+
__+2_____
11) The oxidation number of an element changes from +7 to +2.
This element has (
This represents (
gained
oxidation
lost
) electrons
reduction
).
c) H in LiH _-1__