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Osukunni Practice Questions with Answers - Analytical Chemistry | CHEM 3811, Assignments of Analytical Chemistry

Clayton State UniversityAnalytical Chemistry
Prof. Augustine O. Agyeman

Material Type: Assignment; Professor: Agyeman; Class: Analytical Chemistry; Subject: Chemistry; University: Clayton State University; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/04/2009

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Osukuuni Practice Questions CHEM 3811
1. What is Kb for the equilibrium CN- + H2O ↔ HCN + OH-
(Ka of HCN = 2.1 x 10-9)
Kb = Kw/Ka = 1.0 x 10-14/2.1 x 10-9 = 4.8 x 10-6
2. Calculate the hydronium ion concentration in an aqueous 0.120 M nitrous acid
solution. The principal equilibrium is HNO2 + H2O ↔ H3O+ + NO2-
(Ka = 5.1 x 10-4).
Ka = x2/F
Ka = 5.1 x 10-4 and F = 0.120 M, solve for x = 7.8 x 10-3 M
Percent dissociation = x/F = 6.5%
Percent dissociation is greater than 5% so approximation is not appropriate
Ka = x2/(F-x), solve quadratic equation to get x = [H3O+] = 7.6 x 10-3 M
3. Write the base hydrolysis reaction and calculate the Kb of hypochlorite (OCl-), given
that Ka for hypochlorous acid (HOCl) is 3.0 x 10-8.
OCl- + H2O ↔ HOCl + OH-
Kb = Kw/Ka = 1.0 x 10-14/3.0 x 10-8 = 3.3 x 10-7
4. Calculate the hydronium ion concentration in a solution that is 2.0 x 10-4 M in aniline
hydrochloride, C6H5NH3Cl (Kb = 3.94 x 10-10).
Dissociation of the salt in aqueous solution is complete.
C6H5NH3Cl → C6H5NH3+ + Cl-
C6H5NH3+ + H2O ↔ C6H5NH2 + H3O+ (Ka)
C6H5NH2 + H2O ↔ C6H5NH3+ + OH- (Kb)
Ka = Kw/Kb = 1.00 x 10-14/3.94 x 10-10 = 2.54 x 10-5
Ka = x2/F, solve for x = [H3O+] = 7.1 x 10-5 M
Percent dissociation = x/F = 36%
Percent dissociation is far greater than 5% so approximation is not appropriate
Ka = x2/(F-x), solve quadratic equation to get x = [H3O+] = 5.7 x 10-5 M
5. Calculate the hydronium ion concentration in a 0.075 M NH3 solution. The
predominant equilibrium is NH3 + H2O ↔ NH4+ + OH-
Kb = x2/F, solve for x = [OH-] = 1.2 x 10-3 M
Percent association = x/F = 1.6%, good approximation
[H3O+] = 1.0 x 10-14/1.2 x 10-3 = 8.3 x 10-12 M
pf2

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Osukuuni Practice Questions CHEM 3811

  1. What is Kb for the equilibrium CN-^ + H 2 O ↔ HCN + OH- (Ka of HCN = 2.1 x 10-9) Kb = Kw/Ka = 1.0 x 10-14/2.1 x 10-9^ = 4.8 x 10-
  2. Calculate the hydronium ion concentration in an aqueous 0.120 M nitrous acid solution. The principal equilibrium is HNO 2 + H 2 O ↔ H 3 O+^ + NO 2 - (Ka = 5.1 x 10-4). Ka = x^2 /F Ka = 5.1 x 10-4^ and F = 0.120 M, solve for x = 7.8 x 10-3^ M Percent dissociation = x/F = 6.5% Percent dissociation is greater than 5% so approximation is not appropriate Ka = x^2 /(F-x), solve quadratic equation to get x = [H 3 O+] = 7.6 x 10-3^ M
  3. Write the base hydrolysis reaction and calculate the Kb of hypochlorite (OCl-), given that Ka for hypochlorous acid (HOCl) is 3.0 x 10-8. OCl-^ + H 2 O ↔ HOCl + OH- Kb = Kw/Ka = 1.0 x 10-14/3.0 x 10-8^ = 3.3 x 10-
  4. Calculate the hydronium ion concentration in a solution that is 2.0 x 10-4^ M in aniline hydrochloride, C 6 H 5 NH 3 Cl (Kb = 3.94 x 10-10). Dissociation of the salt in aqueous solution is complete. C 6 H 5 NH 3 Cl → C 6 H 5 NH 3 +^ + Cl- C 6 H 5 NH 3 +^ + H 2 O ↔ C 6 H 5 NH 2 + H 3 O+^ (Ka) C 6 H 5 NH 2 + H 2 O ↔ C 6 H 5 NH 3 +^ + OH-^ (Kb) Ka = Kw/Kb = 1.00 x 10-14/3.94 x 10-10^ = 2.54 x 10- Ka = x^2 /F, solve for x = [H 3 O+] = 7.1 x 10-5^ M Percent dissociation = x/F = 36% Percent dissociation is far greater than 5% so approximation is not appropriate Ka = x^2 /(F-x), solve quadratic equation to get x = [H 3 O+] = 5.7 x 10-5^ M
  5. Calculate the hydronium ion concentration in a 0.075 M NH 3 solution. The predominant equilibrium is NH 3 + H 2 O ↔ NH 4 +^ + OH- Kb = x^2 /F, solve for x = [OH-] = 1.2 x 10-3^ M Percent association = x/F = 1.6%, good approximation [H 3 O+] = 1.0 x 10-14/1.2 x 10-3^ = 8.3 x 10-12^ M
  1. Calculate the pH of 3.0 x 10-5^ M Mg(OH) 2 , which completely dissociates in solution. Mg(OH) 2 ↔ Mg2+^ + 2OH- [OH-] = 2 x 3.0 x 10-5^ M = 6.0 x 10-5^ M [H 3 O+] = 1.0 x 10-14/6.0 x 10-5^ = 1.7 x 10-10^ M pH = - log(1.7 x 10-10) = 9.
  2. Calculate the pH and fraction of dissociation of a 0.0100 M solution of the weak acid with Ka = 1.00 x 10-4. HA + H 2 O ↔ A-^ + H 3 O+ Ka = x^2 /F, solve for x = [H 3 O+] = 1.00 x 10-3^ M Percent dissociation = x/F = 10% Percent dissociation is greater than 5% so approximation is not appropriate Ka = x^2 /(F-x), solve quadratic equation to get x = [H 3 O+] = 9.99 x 10-4^ M pH = - log(9.99 x 10-4) = 3. Fraction of dissociation = x/F = 9.99 x 10-4/0.0100 = 0.
  3. Calculate the pH and fraction of association of a 0.0250 M solution of the weak base with Kb = 1.00 x 10-10. B + H 2 O ↔ BH+^ + OH- Kb = x^2 /F, solve for x = [OH-] = 1.58 x 10-6^ M Percent association = x/F = 6.32 x 10-3%, good approximation (far less than 5%) [H 3 O+] = 1.00 x 10-14/1.58 x 10-6^ = 6.33 x 10-9^ M pH = - log(6.33 x 10-9) = 8. OR pOH = - log(1.58 x 10-6) = 5. pH = 14 - pOH = 14 - 5.801 = 8.