CEE 370

Fall 2010

Homework #1

1. 1 point Problem 2-5 in the Davis & Masten (D&M) text. A solution of sodium bicarbonate is prepared by adding 45.00 g of sodium bicarbonate to a 1.00 L volumetric flask and adding distilled water until it reaches the 1.00 L mark. What is the concentration of sodium bicarbonate in units of (a) milligrams per liter, (b) molarity, (c) normality, and (d) milligrams per liter as CaCO₃?

Given: 45.000 g of sodium bicarbonate in 1.00 L of water.

Solution:

a. The resulting concentration of the solution is 45 g/L. Converting this to mg/L

(45 g/L)(1000 mg/g) = 45,000 mg/L = 4.5 x 10⁴ mg/L

b. To find the molarity, the molecular weight of NaHCO₃ must be found

Na = 22.99 x 1 = 22.99

H = 1.008 x 1 = 1.008

C = 12.01 x 1 = 12.01

3O = 16.00 x 3 = 48.00

Σ = 84.01 g/mol

Then, the molarity

(45 g/L)(1 mol/84 g) = 0.536 M = 0.54 M

c. The equivalent weight of NaHCO₃ is its GMW divided by the number of hydrogen ions transferred. In this case n = 1 because Na⁺ is replaced by 1 H. Thus normality (N) is n*M, and in this case n = 1

(0.536 M)(1 N/M) = 0.536 M = 0.54 N

d. Change to units of mg/L as CaCO₃, using Eqn 2-87

50 mg/L as CaCO₃

45,000 mg/L *(-------------------------) = 26,785 mg/L = 2.7 x 10⁴ mg/L as CaCO₃

84 mg/L as NaCO₃

2. Problem 2-6 in the D&M text. Balance the following 5 chemical equations:

1 point

Solution:

a. CaCl₂ + Na₂CO₃ = CaCO₃ + NaCl

Elements Reactants Products

Ca 1 1

Cl 2 1

Na 2 1

C 1 1

O 3 3

Note that we are short 1 Na, so multiply product NaCl by 2

Elements Reactants Products

Ca 1 1

Cl 2 2

Na 2 2

C 1 1

O 3 3

This yields the balanced equation:

CaCl₂ + Na₂CO₃ = CaCO₃ + 2 NaCl

b. C₆H₁₂O₆ + O₂ = CO₂ + H₂O

Elements Reactants Products

C 6 1

H 12 2

O 8 3

Note that we are short 5 C, 10 H and 5 O. Multiply product CO₂ by 6 and H₂O by 6. This balances C and H on each side, and leaves a difference of 10 O. Multiply the reactant O₂ to balance equation:

C₆H₁₂O₆ + 6 O₂ = 6 CO₂ + 6 H₂O

Elements Reactants Products

C 6 6

H 12 12

O 18 18

c. NO₂ + H₂O = HNO₃ + NO

Elements Reactants Products

N 1 2

H 2 1

O 3 4

Note that H is short 1, therefore multiply the product HNO₃ by 2. This causes N to be out of balance, and the reactant NO₂ should be multiplied by 3.

3 NO₂ + H₂O = 2 HNO₃ + NO

Elements Reactants Products

N 3 3

H 2 2

O 7 7

d. C₄H₁₀ + O₂ = CO₂ + H₂O

Elements Reactants Products

C 4 1

H 10 2

O 2 3

Note that H is out of balance, but in order to keep the number of O even, we must multiply the product H₂O by 10. As a result, the reactant C₄H₁₀ must be multiplied by 2.

2 C₄H₁₀ + O₂ = CO₂ + 10 H₂O

Elements Reactants Products

C 8 1

H 20 20

O 2 12

However, the reaction is still not balanced. CO₂ must be multiplied by 8 to maintain an equal number of C. As a result, O is left unbalanced. Multiply O₂ by 13.

2 C₄H₁₀ + 13 O₂ = 8 CO₂ + 10 H₂O

Elements Reactants Products

C 8 8

H 20 20

O 26 26

e. Al(OH)₃= Al³⁺ + OH^-

Elements Reactants Products

Al 1 1

O 3 1

H 3 1

Note that Al is in balance, however we are missing two OH groups. Multiply the product OH^- by 3.

Al(OH)₃= Al³⁺ + 3 OH^-

Elements Reactants Products

Al 1 1

O 3 3

H 3 3

3. 1 point You've just prepared a solution by dissolving 20 mg sodium sulfide (Na2S), and 30 mg potassium sulfate dihydrate (K2SO42H2O) in 1 Liter of distilled water.

a) What is the molar concentration of sodium sulfide in this solution?

b) What is the equivalent concentration of sodium sulfide in this solution?

c) What is the molar concentration of potassium in this solution?

d) What is the concentration of total sulfur in this solution in mg/L?

e) What is the concentration of reduced sulfur (i.e., S(-II)) in this solution in mg/L?

f) What is the theoretical TDS of this solution in mg/L?

Answers:

Assemble essential data:

	Atomic Wt
Na	22.99
S	32.06
K	39.1
O	16
H	1.01

Calculate molar quantities and percentages

		Amount Added
GFW		mg	mMoles	%water	%Sulfur	%S(-II)
78.04	Na₂S	20	0.256279	0.00%	41.08%	41.08%
210.3	K₂SO₄ 2H₂O	30	0.142653	17.14%	15.24%	0.00%

a. Molar Concentration

GFW = 2* 22.99 + 32.06 = 78.04

0.26mM

b. Equivalent Concentration

Since Z=2 for the compound (i.e., there are 2 positive charges Na⁺ and Na⁺, and 2 negative charges S^-2), then GEW = GFW/2 = 39.02, This also makes the equivalent concentration twice the molar concentration, so:

0.51 meq/L

c. Molar Concentration of K

first determine the molar concentration of the potassium sulfate

GFW = 2* 39.1 + 32.06 + 4*16 + 2*(16+2*1.10) = 210.3

0.143mM

Then translate this to a concentration of potassium alone

0.29mM

d. Concentration of Total Sulfur

One approach is to determine total molar sulfur concentration and then convert to mg/L units.

= 12.8 mg/L

e. Concentration of Reduced Sulfur

The reduced sulfur is only the S in Na₂S. This is sulfur in its (-II) oxidation state.

= 8.2 mg/L

f. What is the TDS

Determine the total mass of salts added, minus any water (i.e., water of crystallization).

TDS = 44.9 mg/L

4. 0.5 points Problem 2-12 in the D&M text. If 200 mg of HCl is added to water to achieve a final volume of 1.00 L, what is the final pH?

pH of HCl solution

Given: 200 mg of HCl in 1.00 L

Solution:

a. Calculate molarity

GMW of HCl = 36.45

200 mg g 1

(------------)(10^-3 ------)(-----------------) = 5.487 x 10^-3 M of HCl

L mg 36.45 g/mol

b. Moles of H⁺ on ionization

HCl ↔ H⁺ + Cl^-

so 1 mole HCl = 1 mole H⁺

[H⁺] = 5.476 x 10^-3 M

c. Using Eqn 2-38

pH = - log(5.487 x 10^-3) = 2.26

5. 0.5 points What amount (mass, in mg) of NaOH (a strong base), would be required to neutralize the acid in Problem 2-12? (see Problem 2-29 & Example 2-11)

From the problem above, we now know that 200 mg/L of HCl is 5.476 x 10^-3 M

Since HCl has a single hydrogen ion it donates to water, and NaOH has a single hydroxide that can neutralize a single hydrogen ion, the stoichiometry of neutralization is 1:1

Therefore we need 5.476 x 10^-3 M of NaOH

The GMW of NaOH is 22.99 + 16 + 1.008 = 40 g/M

So the amount of NaOH is 40x 5.476 x 10^-3 M = 0.219 g

6. 1 point Problem 2-15 in the D&M text. The concentration of a chemical degrades in water according to first-order kinetics. The degradation constant is 0.2 day^-1. If the initial concentration is 100.0 mg/L, how many days are required for the concentration to reach 0.14 mg/L? Also calculate the half-life (t_1/2) for this decay reaction.

Given: 1^st order kinetics, rate of 0.2 d^-1, initial concentration = 100 mg/L.

Solution:

a. Time to read 0.14 mg/L by 1^st order kinetics

ln ---- = - kt

C_o

0.14

ln -------- = - (0.2)t

100

-6.57 = -0.2t

t = 32.85 or 32.9 d

b. Half Life

=3.5 d

7. not
graded Problem 2-28 in the D&M text. Each mole of CaF₂(s) dissolved yields 1 mole of Ca²⁺ and 2 moles of F^- (fluoride). The solubility product of calcium fluoride (CaF₂) is 3x10^-11 at 25C. Could a fluoride concentration of 1.0 mg/L be obtained in water that contains 200 mg/L of calcium? Show your work.

Given: Solubility product of CaF₂ = 3 x 10^‑11 , F = 1.0 mg/L and Ca = 200 mg/L

Solution:

a. Convert Ca and F to moles/L

200 mg/L

Ca = ‑‑--------‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = 4.99 x 10^‑3 moles/L

(40.08 g/mole)(1000 mg/g)

1.0 mg/L

F = ‑‑‑‑‑‑‑‑‑‑---------‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ = 5.26 x 10^‑5 moles/L

(18.998 g/mole)(1000 mg/g)

b. Calculate solubility of F with 200 mg/L of Ca in solution.

K_s = [Ca][F]²

K_s = [4.99 x 10^‑3][F]² = 3.00 x 10^‑11

3.00 x 10^‑11

[F] = (‑‑‑----‑‑‑‑‑‑‑‑‑‑‑‑)^1/2 = 7.75 x 10^‑5 moles/L

4.99 x 10^‑3

c. Since 7.75 x 10^‑5 is greater than 5.26 x 10^‑5, the 1.0 mg/L of F will be soluble.

Assigned: 10 Sept 10

Homework #1

Answers:

a. Molar Concentration

b. Equivalent Concentration

c. Molar Concentration of K

d. Concentration of Total Sulfur

e. Concentration of Reduced Sulfur

f. What is the TDS

Due: 20 Sept 10