CHEM 10050
Problem Set #3
Name_________________
1. Calculate the molarity of normal saline (0.90% w/w NaCl) given the density of this
solution to be 1.05 g/ml.
Assume 100 g solution.
0.90% 
Then:
100 g soln x
x g NaCl
x 100%
100 g soln
1ml soln
1L
 95.2 ml soln x
 0.0952 L
1.05 g
1000 ml
0.90 g NaCl x
1mol NaCl
 0.015 mol NaCl
58.4 g NaCl
0.015 mol NaCl
 0.16 M
0.0952 L
2. Calculate the molarity of an aqueous solution resulting when 275 ml of 3.15 M
glucose is diluted with 150 ml of water.
M1V1 = M2V2
Total volume = 275 ml + 150 ml = 425 ml
3.15 M x 275 ml = M2 x 425 ml
M2 
3.15 M x 275 ml
 2.04 M
425 ml
3. How many grams of sodium hydroxide are in 36.7 ml of 2.256 M sodium hydroxide?
36.7 ml x
1L
2.256 mol 40.0 g NaOH
x
x
 3.31 g NaOH
1000 ml
1L
1 mol NaOH
4. What is the concentration of glucose in ppm in a 0.00025 M glucose solution? The
density of this solution is 1.02 g/ml.
0.00025 mol glucose
1L soln
0.00025 M 
0.00025 mol glucose x
1L x
180 g glucose
 0.045 g glucose
1mol glucose
1000 ml 1.02 g
x
 1.02 x 10 3 g solution
1L
1 ml
ppm 
0.045 g
x 10 6  44 ppm
1.02 x 10 3 g
5. Draw reaction diagrams for endothermic and exothermic reactions. Label each axis,
reactants, products, heat of reaction (H), and energy of activation (Ea). Define
heat of reaction and energy of activation.
Endothermic Reaction
A + B C
Exothermic Reaction
A + B C
‡
AB
C
products
Energy
Ea
ΔH
A+B
reactants
A+B
reactants
Energy
‡
AB
Ea
ΔH
Reaction Pathway
C
products
Reaction Pathway
6. Given the following; [H2] = 0.00537, [HCl] = 0.0354, H = -56.3 kJ, G = +26.6 kJ;
Calculate Keq for the following reaction:
Al(s) + HCl(g) 
 H2(g) + AlCl3(s)
Balanced equation:
2Al(s) + 6HCl(g) 
 3H2(g) + 2AlCl3(s)
K eq 
[H2 ]3
[0.00537]3

 78.7
[HCl]6
[0.0354] 6
a) Are the reactants or products favored at equilibrium? Explain.
Products are favored at equilibrium. K is larger than 1.
b) Is this reaction exergonic or endergonic? Explain.
The reaction is endergonic (non-spontaneous). G = (+)
c) Is S positive or negative for this reaction? Explain your answer.
S = (-). Eight (8) moles of reactants is producing five (5) moles of products. The
system is becoming more organized.
d) Predict the effects of the following on the equilibrium and K (eq):
i.
increase [H2]
Equilibrium shifts left. Keq is constant.
ii. decrease [HCl]
Equilibrium shifts left. Keq is constant.
iii. decrease pressure
Equilibrium shifts left. Keq is constant.
iv. increase temperature Equilibrium shifts left. Keq decreases.
v. add a catalyst
No effect.
7. Calculate the pH, [H+], and [OH-] of a 0.0256 M solution of hydrobromic acid.
HBr is a strong acid (electrolyte): HBr  H+ + Br[H+] = 0.0256 M
pH = -log[H+] = -log[0.0256] = 1.59
pOH = 14 – 1.59 = 12.41
[OH-] = 10-12.41 = 3.89 x 10-13 M
8. What is the pH of an acetic acid/sodium acetate buffer solution that has the
following solution concentrations [HOAc] = 0.100 M, [OAc-] = 0.00500 M. For HOAc
Ka = 1.8 x 10-5.
[A  ]
pH of a buffer solution: pH = pKa + log
[HA ]
[0.00500]
pH = -log[1.8 x 10-5] + log
[0.100]
pH = 4.74 + (-1.30) = 3.44
9. Predict the products for the following acid-base reactions. Label the conjugate pairs
and balance each equation.
a) Na2CO3 + 2HCl  2NaCl + [H2CO3]  CO2 + H2O
base1
acid1
base2
acid2
b) NH3 + H2O  NH4+ + OHbase1
acid1
acid2
base2
c) CN- + H2SO4  HCN + HSO4base1
acid1
acid2
base2
d) HPO42- + OH-  H2O + PO43acid1
base1
acid2 base2
e) NH3 + HBr  NH4+ + Brbase1
acid1
acid2
base2
f) HCO3- + H2O  H2CO3 + OHbase1
acid1
acid2
base2
or
HCO3- + H2O  CO32- + H3O+
acid1
base1 base2
acid2
g) 2HF + Ca(OH)2  CaF2 + 2H2O
acid1
base1
base2
acid2
10. Explain in detail how the carbonic acid/bicarbonate buffer system in your body
regulates blood pH. Be certain to use Lé Châtliér’s Principle and chemical
equations in your explanation.
See notes.