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

Name ________________________

(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 the sample?

nitrogen

(6 pts.) 5. Give the mass balance equation for H₃PO₄ in an acid-base titration.

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

(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 strong 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

9. Derive the titration curve for the titration of 45.00 mL of 0.1100 M HCl with 0.1100 M NaOH.

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

initial point

[H⁺] = C_HCl = 0.1100 M

pH = - log[H⁺] = - log(0.1100) = 0.9586

(4 pts.) (b.) At 25.00 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⁺] = (45.00mL*0.1100M) - (25.00mL*.1100M)/(45.00 + 25.00)mL = 3.142x10^-2M

pH = - log[H⁺] = - log(3.142x10^-2) = 1.5027

(2 pts.) (c.) At 45.00 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 80.00 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^-] = (80.00mL*0.1100M) - (45.00mL*.1100M)/(45.00 + 80.00)mL = 3.080x10^-2M

pOH = - log[OH^-] = - log(3.080x10^-2) = 1.5114

pH = 14.00 -pH = 14.00 - 1.5114 = 12.4886

(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 formic acid and sodium formate.

3.7

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

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

Fe(OH)_3(s) <=> Fe⁺³ + 3(OH)^-

2 H₂O <=> H₃O⁺ + OH^-

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

let x = molar solubility = [Fe⁺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 = [Fe⁺³][(OH)^-]³ = 1.6 x 10^-39M⁴

K_w = [H₃O⁺][OH^-] = 1.00 x 10^-14M²

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

[OH^-] = 3 [Fe⁺³] + [H₃O⁺]

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

[OH^-] = 3 [Fe⁺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^-] ~ [Fe⁺³]

3 [Fe⁺³] >> [H₃O⁺]

Step 8. Solve algebraic equations.

then K_sp = [Fe⁺³][(OH)^-]³ = 1.6 x 10^-39M⁴

K_sp = [Fe⁺³][3Fe⁺³]³ = 1.6 x 10^-39M⁴

27 [Fe⁺³]⁴ = 1.6 x 10^-39M⁴

[Fe⁺³] = 8.7 x 10^-11M ~ [(OH)^-]

Step 9. Check validity of assumptions.

3 [Fe⁺³] >> [H₃O⁺]

[H₃O⁺] = K_w/[(OH)^-]

[H₃O⁺] = (1.00 x 10^-14M²/7.4 x 10^-12M) = 0.00135 M

assumption invalid

Solve simultaneous equations

[OH^-] = 3[Fe⁺³] + [H₃O⁺]

[H₃O⁺] = K_w/[(OH)^-]

[OH^-] = 3[Fe⁺³] + K_w/[(OH)^-]

[OH^-] - K_w/[(OH)^-] = 3[Fe⁺³]

([OH^-]/3) - (K_w/3[(OH)^-]) =[Fe⁺³]

K_sp = [Fe⁺³][(OH)^-]³ = 1.6 x 10^-39M⁴

([OH^-]/3) - (K_w/3[(OH)^-])[(OH)^-]³ = 1.6 x 10^-39M⁴

([OH^-]⁴/3) - (K_w*[(OH)^-]²/3) = 1.6 x 10^-39M⁴

[OH^-]⁴ - (K_w*[(OH)^-]² = 3*1.6 x 10^-39M⁴

[OH^-]⁴ - (1.00 x 10^-14M²*[(OH)^-]² = 3*1.6 x 10^-39M⁴

[OH^-]⁴ - (1.00 x 10^-14M²*[(OH)^-]² = 4.8 x 10^-39M⁴

method of iterations

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

[OH^-]⁴ - (1.00 x 10^-14M²*[(OH)^-]² = 4.8 x 10^-39M⁴

6.5610000000000E-29 - (1.00E-14)(8.1000000000000E-15) = 4.80E-39

6.561000000000E-29 - 8.1000000000000E-29 = 4.80E-39

-1.54E-29 = 4.80E-39

assumed value too small because left side of equation is less than the right

assume a larger value: [OH^-] = 1.00e-7 M

0.00E+00 = 4.80E-39

assumed value still too small but much closer, left side still less than the right

assume a slightly larger value: assume [OH^-] = 1.1e-7 M

2.54E-29 = 4.80E-39

assumed value still too large because left side of the equation is now larger than the right

after several more tries: assume [OH^-] = 1.000000000024E-07 M

4.80E-29 = 4.80E-39

assumed value is an exact solution

thus

[Fe⁺³] = (K_sp/[OH^-]³) = (1.60E-39 M⁴/1e-21 M³) = 1.6E-18 M

(3 pts.) 13. Draw the first derivative curve for the equivalence point of a strong acid-strong base titration.

(3 pts.) 14. Draw the second derivative curve for the equivalence point of a strong acid-strong base titration.

15. Derive the titration curve for the titration of 45.00 mL of 0.1100 M aminobenzene (aniline) with 0.1100 M NaOH.

Aminobenzene (aniline) is a weak acid with K_a = 2.51 x 10^-5M. This value was obtained from Appendix G, page AP15.

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

[H⁺] = (K_a*C_HA)^1/2

[H⁺] = (2.51 x 10^-5M * 0.1100 M)^1/2 = 1.66 x 10^-3M

pH = 2.78

(8 pts.) (b.) At 25.00 mL of NaOH added<b>
V_acidM_acid > V_baseM_base, buffer region

[Acid]_excess = ((V_acidM_acid - V_baseM_base)/(V_acid + V_base))

[Acid]_excess = ((45.0 mL)(0.1100 M) - (25.0 mL)(0.1100 M))/((45.0 + 25.0)mL) = 3.14 x 10^-2M

[conjugate base] = ((V_baseM_base)/ (V_acid + V_base))

[conjugate base] = ((25.0 mL)(0.1100 M))/((45.0 + 25.0)mL) = 3.93 x 10^-2M

[H⁺] = K_a*([Acid]/[Conjugate base])

[H⁺] = (2.51 x 10^-5M)*(3.14 x 10^-2M/3.93 x 10^-2M) = 2.01 x 10^-5M

pH = 4.70

(6 pts.) (c.) At 45.00 mL of NaOH added

V_acidM_acid = V_baseM_base, equivalence point
K_h = K_w/K_a = (1.00 x 10^-14/2.51 x 10^-5)M = 3.98 x 10^-10M

[Conjugate base] = ((V_acidM_acid)/(V_acid + V_base))

[Conjugate base] = ((45.0 mL)(0.1100 M))/((45.0 + 45.0)mL) = 0.0550 M

[OH^-] = (K_h*[Conjugate base])^1/2

[OH^-] = ((3.98 x 10^-10M )*0.0550 M)^1/2 = 4.68 x 10^-6M

pOH = 5.33

pH = 14.00 + pOH = 8.67

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

phenolphthalein

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: ___________________