In-Class Section         ______
Take-Home Section   ______
Combined Score         ______

Name ________________________

Quantitative Analysis
Exam 2
In-Class Section
March 22, 2005

(4 pts.) 1. Describe the difference between the end-point and the equivalence-point of a titration.

The end-point is when a physical change, such as change in color of indicator, takes place. This can indicate the equivalence point of a titration if the correct indicator is selected.

The equivalence point is the condition in a titration where the amount of analyte equals, reactivity wise, the amount of titrant added. Determining this point is the point of the titration.

(5 pts.)    2. Draw a typical strong base-strong acid titration curve, labeling the axes and the equivalence point.  (HINT: The strong base is the analyte and the strong acid is the titrant!)

(7 pts.)    3. Draw a typical diprotic weak acid-strong base titration curve, labeling the axes, the equivalence points and the buffer regions.

(6 pts.)    4. Give the charge balance equation for carbonic acid, H₂CO₃, in a diprotic weak acid-strong base titration.

C_T = [H₂CO₃] + [HCO₃^-1] + [CO₃^-2]

(4 pts.)    5. List two characteristics of a substance which can be used as a primary standard.

Any two of the following:
• High purity
• Stability toward air
• Absence of hydrate water
• Available at moderate cost
• Soluble
• Large F.W.

(4 pts.)    6. Describe what constitutes a buffer solution.

A buffer solution can be made from a weak acid and its conjugate base,
or
from a weak base and its conjugate acid.

(2 pts.)    7. The Kjeldahl Analysis is used to determine the amount of which element found in a sample?

nitrogen

(4 pts.)    8. Explain why phenolphthalein indicator is never used in a strong base-strong acid titration but can be used in a strong acid-strong base titration (First listed substance is the analyte and the second listed substance is the titrant.

In a titration involving a strong base being titrated with a strong acid, the red color would appear first. going to colorless. The last trace of pink color is extremely hard for a person to see coming from red.

(6 pts.)    9. Give the mass balance equation for phosphoric acid, H₃PO₄, in a diprotic weak acid-strong base titration.

C_T = [H₃PO₄] + [H₂PO₄^-1] + [HPO₄^-2] + [PO₄^-3]

Quantitative Analysis
Exam 2
Take-Home Section
Due 12:00noon on March 28, 2005

All work must appear on the exam!

10. Derive the titration curve for the titration of 40.00 mL of 0.100 M HCl with 0.200 M NaOH.

(3 pts.)            (a.) At 0.00 mL of NaOH added

initial point

M_HCl = [H⁺] = 0.100 M

pH = - log ([H⁺]) = 1.000

(4 pts.)            (b.) At 15.00 mL of NaOH added

V_acidM_acid> V_baseM_base, pre-equivalence point

[H⁺] = (V_a*M_a) - (V_b*M_b)/(V_a + V_b)

[H⁺] = (40.00mL*0.1000M) - (15.00mL*.2000M)/(40.00 + 15.00)mL = 1.818x10^-2M

pH = - log[H⁺] = - log(1.818x10^-2) = 1.7404

(2 pts.)             (c.) At 20.00 mL of NaOH added

V_acidM_acid = V_baseM_base, equivalence point

for the equivalence point of a strong acid - strong base titration, the pH is 7.00.

(5 pts.)             (d.) At 40.00 mL of NaOH added

V_baseM_base > V_acidM_acid, pre-equivalence point

[OH^-] = (V_b*M_b) - (V_a*M_a)/(V_a + V_b)

[OH^-] = (40.00mL*0.1000M) - (40.00mL*.2000M)/(40.00 + 40.00)mL = 5.000x10^-2M

pOH = - log[OH^-] = - log(5.000x10^-2) = 1.3010

pH = 14.00 -pH = 14.00 - 1.3010 = 12.70

(2 pts.)    11. Select an appropriate indicator for the titration in Problem 10.

phenolphthalein

(2 pts.)    12. Give an appropriate pH for a buffer solution composed of benzoic acid and sodium benzoate.

pK_a = 4.202 for benzoic acid; each molar amounts of benzoic acid and sodium benzoate

(18 pts.)  13. Using the 9 step procedure, calculate solubility of NiS(b) in water.

Step 1. Write a set of balanced chemical equations for all pertinent equilibria.

NiS --> Ni⁺² + S^-2

S^-2 + H₂O ---> HS^- + OH^-

HS^- + H₂O ---> H₂S + OH^-

H₂O ---> H⁺ + OH^-

Step 2. State in terms of equilibrium concentrations what quantity is being sought.

let x = molar solubility = [Ni⁺²]

Step 3. Write equilibrium-constant expressions for all equilibria developed in Step 1, and find numerical values for the constants in tables of equilibrium constants.

NiS --> Ni⁺² + S^-2
K_sp = [Ni⁺²][S^-2] = 4 x 10^-20

S^-2 + H₂O ---> HS^- + OH^-
K_b1 = [HS^-][OH^-]/[S^-2] = 0.80

HS^- + H₂O ---> H₂S + OH^-
K_b2 = [H₂S][OH^-]/[HS^-] = 1.1x10^-7

H₂O ---> H⁺ + OH^-
K_w = [H⁺][OH^-] = 1.0x10^-14

Step 4. Write mass-balance expressions for the system.

C_T = [H₂S] + [HS^-] + [S^-2] = [Ni⁺²]

Step 5. If possible, write a charge balance expression for the system.

2[Ni⁺²] + [H⁺] = [HS^-] + 2[S^-2] + [OH^-]

Step 6. Count equations vs. unknowns. If more unknowns than equations, seek additional equations, or make appropriate approximations.

6 equations and 6 unknowns

Step 7. Make suitable approximations to simplify the algebra.

None used

Step 8. Solve algebraic equations.

Solved with Spreedsheet

[Ni⁺²] = 3.16x10^-7M

Step 9. Check validity of assumptions.

Not applicable because no approximations made

14. Derive the titration curve for the titration of 50.00 mL of 0.1200 M pentanoic acid with 0.1200 M NaOH.

K_a = 1.44x10^-5

(5 pts.)             (a.) At 0.00 mL of NaOH added

H⁺] = (K_a*C_HA)^1/2 = (1.44E-05*0.1200)^1/2 = 0.00131M

pH = - log[H⁺] = - log (0.00131) = 2.881

(8 pts.)             (b.) At 25.00 mL of NaOH added

V_acidM_acid>V_NaOHM_NaOH, thus, buffer region

[acid]_excess = (V_acidM_acid - V_NaOHM_NaOH)/(V_acid + V_NaOH)

= ((50.00mL *0.1200M) - (25.00mL* 0.1200M))/(50.00 + 25.00)mL = 0.04000M

[salt] = (V_NaOHM_NaOH)/(V_acid + V_NaOH)

= (25.00mL* 0.1200M)/(50.00 + 25.00)mL = 0.04000M

[H⁺] = K_a*[acid]_excess /[salt] = (1.44E-05*0.04000/0.04000) = 1.44E-05M

pH = - log[H⁺] = - log (1.44e-5) = 4.84

(6 pts.)             (c.) At 50.00 mL of NaOH added

V_acidM_acid = V_NaOHM_NaOH, thus equivalence point

[salt] = (V_NaOHM_NaOH)/(V_acid + V_NaOH)

[salt] = (50.00mL*0.1200M)/(50.00 + 50.00)mL = 0.06000M

[OH^-] = ((K_w/K_a)[salt])^1/2

[OH^-] = ((1.00e-14/1.44e-5)(0.06000))^1/2 = 6.49e-6M

pOH = - log([OH^-]) = - log(6.14e-6) = 5.19

pH = 14.00 - pOH = 14.00 - 5.19 = 8.81

(2 pts.)    15. Select an appropriate indicator for the titration in Problem 14.

phenolphthalein

(3 pts.)    16. What is the pH of a 1.55x10^-8 M solution of HCl?

[H⁺]_total = [H⁺]_water + [H⁺]_HCl

[H⁺]_HCl = 1.55x10^-8M

let [H⁺]_water = x = [OH^-]

K_w = [H⁺]_total[OH^-] = 1.01x10^-14M² at 25^oC

[H⁺]_total[OH^-] = 1.01x10^-14M²

([H⁺]_water + 1.55x10^-8M) * [OH^--] = 1.01x10^-14M²

( x + 1.55x10^-8M) * x = 1.01x10^-14M²

by quadratic: x = 9.30x10^-8M

[H⁺]_total = 9.30x10^-8M + 1.55x10^-8M

[H⁺]_total = 10.85x10^-8M = 1.09x10^-7M

pH = 6.96