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

Quantitative Analysis
Exam 2
In-Class Section
March 27, 2003

Under Construction

(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.

(4 pts.) 2. 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.) 3. 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.) 4. The Kjeldahl Analysis is used to determine the amount of which element found in a sample?

nitrogen

(6 pts.) 5. Give the mass balance equation for H₂CO₃ in an acid-base titration.

C_T = [H₂CO₃] + [HCO₃^-1] + [CO₃^-2]

(5 pts.) 6. Give the charge balance equation for the following equilibria:

HgS_(s) ó Hg⁺² + S^-2

S^-2 + H₂O ó HS^- + OH^-

HS^- + H₂O ó H₂S + OH^-

H₂O ó H⁺ + OH^-

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

(6 pts.) 7. Draw a typical weak acid-strong base titration curve, labeling the axes and the equivalence point.

(9 pts.) 8. Draw a typical diprotic weak acid-strong base titration curve, labeling the axes, the equivalence points and the buffer regions.

Name

Quantitative Analysis
Exam 2
Take-Home Section
March 27, 2003

9. Derive the titration curve for the titration of 25.00 mL of 0.0990 M HCl with 0.1100 M NaOH.

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

pH = -log[HCl] = -log(0.0990) =1.0044

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

[H+] = (V_HClM_HCl - V_NaOHM_NaOH)/(V_HCl + V_NaOH)

= ((25.00mL *0.0990M) - (15.00mL* 0.1100M))/(25.00 + 15.00)mL = 0.020625M

pH = - log[H⁺] = - log(0.02063) = 1.6855

(2 pts.) (c.) At 22.50 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

[OH^-] = ((V_NaOHM_NaOH) - (V_HClM_HCl))/(V_HCl + V_NaOH)

= ((40.00mL*0.1100M) - (25.00mL*0.0990M))/(25.00 + 40.00)mL = 0.02962M

pOH = - log[OH^-] = - log(0.02962) =1.5284

pH = 14.00 - pOH = 14.00 - 1.5284 = 12.47

(2 pts.) 10. Select an appropriate indicator for the titration in Problem 9.

phenolphthalein

(2 pts.) 11. Give an appropriate pH for a buffer solution composed of butanoic acid and sodium butanoate.

for butanoic acid: K_a = 1.52x10^-5M

the most appropriate pH for a buffer solution is the pK_a

pK_a = - log(K_a) = - log (1.52x10^-5) = 4.819

(18 pts.) 12. Using the 9 step procedure, calculate solubility of Cr(OH)₃ in water.

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

Cr(OH)_3(s) <=> Cr⁺³ + 3(OH)^-

2 H₂O <=> H₃O⁺ + OH^-

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

let x = molar solubility = [Cr⁺³]


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.

K_sp = [Cr⁺³][(OH)^-]³ = 1.6 x 10^-30M⁴

K_w = [H₃O⁺][OH^-] = 1.00 x 10^-14M²

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

[OH^-] = 3 [Cr⁺³] + [H₃O⁺]

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

[OH^-] = 3 [Cr⁺³] + [H₃O⁺]

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

3 unique equations and 3 unknowns

Step 7. Make suitable approximations to simplify the algebra.

assume [OH^-] ~ 3[Cr⁺³]

3 [Cr⁺³] >> [H₃O⁺]

Step 8. Solve algebraic equations.

then K_sp = [Cr⁺³][(OH)^-]³ = 1.6 x 10^-30M⁴

K_sp = [Cr⁺³][3Cr⁺³]³ = 1.6 x 10^-30M⁴

27 [Cr⁺³]⁴ = 1.6 x 10^-30M⁴

[Cr⁺³] = 1.6 x 10^-8M and [(OH)^-] ~ 3[Cr⁺³] = 3(1.6 x 10^-8M) = 4.8 x 10^-8M

Step 9. Check validity of assumptions.

3 [Cr⁺³] >> [H₃O⁺]

[H₃O⁺] = K_w/[(OH)^-]

[H₃O⁺] = (1.00 x 10^-14M²/4.8 x 10^-8M) = 2.1 x 10^-7M

assumption invalid

Solve simultaneous equations

[OH^-] = 3[Cr⁺³] + [H₃O⁺]

[H₃O⁺] = K_w/[(OH)^-]

[OH^-] = 3[Cr⁺³] + K_w/[(OH)^-]

[OH^-] - K_w/[(OH)^-] = 3[Cr⁺³]

([OH^-]/3) - (K_w/3[(OH)^-]) =[Cr⁺³]

K_sp = [Cr⁺³][(OH)^-]³ = 1.6 x 10^-30M⁴

([OH^-]/3) - (K_w/3[(OH)^-])[(OH)^-]³ = 1.6 x 10^-30M⁴

([OH^-]⁴/3) - (K_w*[(OH)^-]²/3) = 1.6 x 10^-30M⁴

[OH^-]⁴ - (K_w*[(OH)^-]² = 3*1.6 x 10^-30M⁴

[OH^-]⁴ - (1.00 x 10^-14M²*[(OH)^-]² = 3*1.6 x 10^-30M⁴

[OH^-]⁴ - (1.00 x 10^-14M²*[(OH)^-]² = 4.8 x 10^-30M⁴

method of iterations

assume [OH^-] = 9.00E-08 M

[OH^-]⁴ - (1.00 x 10^-14M²*[(OH)^-]² = 4.8 x 10^-30M⁴ using solver on a calculator

thus

[Cr⁺³] = (K_sp/[OH^-]³) = (1.60E-30 M⁴/1e-21 M³) = 1.5E-9 M

13. Derive the titration curve for the titration of 25.00 mL of 0.0990 M 1,10-phenanthroline with 0.1100 M NaOH.

Ka = 1.38E-05

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

[H⁺] = (K_a*C_HA)^1/2 = (1.38E-05*0.0990)^1/2 = 0.00117M

pH = - log[H⁺] = - log (0.00117) = 2.932

(8 pts.) (b.) At 15.00 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)

= ((25.00mL *0.0990M) - (15.00mL* 0.1100M))/(25.00 + 15.00)mL = 0.02063M

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

= (15.00mL* 0.1100M)/(25.00 + 15.00)mL = 0.04125M

[H⁺] = K_a*[acid]_excess /[salt] = (1.38E-05*0.02962/0.04125) = 9.91E-06M

pH = - log[H⁺] = - log (9.91e-6) = 5.00

(6 pts.) (c.) At 22.50 mL of NaOH added

V_acidM_acid = V_NaOHM_NaOH, thus equivalence point

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

[salt] = (22.50mL*0.1100M)/(25.00 + 22.50)mL = 0.05211M

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

[OH^-] = ((1.00e-14/1.38e-5)(0.05211))^1/2 = 6.14e-6M

pOH = - log([OH^-]) = - log(6.14e-6) = 5.21

pH = 14.00 - pOH = 14.00 - 5.21 = 8.78

(7 pts.) 14. What is the pH of a 1.55x10^-8 M solution of NaOH?

It is a solution of NaOH, therefore, its pH must be basic, greater than 7!

[OH^-]_total = [OH^-]_water + [OH^-]_NaOH

[OH^-]_total = [OH^-]_water + 1.55x10^-8M

let x = [OH^-]_water = [H⁺]

K_w = [H⁺][OH^-]_total = 1.00x10^-14M²

K_w = x(x + 1.55x10^-8M) = 1.00x10^-14M²

by quadratic equation

x = 9.3x10^-8M

[OH^-]_total = [OH^-]_water + 1.55x10^-8M

[OH^-]_total = x + 1.55x10^-8M

[OH^-]_total = 9.3x10^-8M + 1.55x10^-8M = 1.09E-07M

pOH = 6.96 thus pH = 7.04

(2 pts.) 15. Draw the second derivative curve for the equivalence point of a weak acid-strong base titration.

(2 pts.) 16. Select an appropriate indicator for each equivalence point for the titration in problem 13.

phenolphthalein or thymol blue

I neither received nor gave any assistance on this examination nor did I see anyone else give or receive assistance on this examination.

Signature: ________________________________________

Date: ___________________