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8 Problems on Aqueous Equilibrium - Worksheet 5 | CH 302, Assignments of Chemistry

The University of Texas at AustinChemistry
Prof. David A. Laude

Material Type: Assignment; Professor: Laude; Class: PRINCIPLES OF CHEMISTRY II; Subject: Chemistry; University: University of Texas - Austin; Term: Fall 2001;

Typology: Assignments

Pre 2010

Uploaded on 08/30/2009

koofers-user-igl
koofers-user-igl 🇺🇸

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Worksheet 5. Aqueous Equilibrium Problems; Simple Equilibria
1. Identify the acid/base and their conjugate base/acid, and which definition you use to determine(Bronsted,
Arrhenius or Lewis):
a. HCO3- + H+ ↔ H2CO3
Base conj acid: Bronsted
b. HCO3- ↔ CO32- + H+
Acid conj base : Arrhenius
c. CH3NH2 + H2O ↔ CH3NH3+ + OH-
Base acid conj acid conj base : Lewis
d. C6H5OH + H2O ↔ C6H5O- + H3O+
Acid base conj base conj acid : Lewis, Arrhenius, Bronsted
e. H2O + H2O ↔ H3O + + OH-
Acid base conj acid conj base
-
2. Assuming Kw = 1x10-14, calculate the molarity of OH- in solutions at 25ºC when the H+ concentration is:
a. 0.2M
At 25ºC, Kw = [OH-] [H+] = 1x10-14
[OH-] = 1x10-14/ .2 = 5x10-14
b. 5 x 10-10 M
[OH-] = 1x10-14/ 5 x 10-10 = 2 x 10-5
c. 100 M
[OH-] = 1x10-14/ 100 = 1x10-16
3. For each of these strong acid/base solutions, calculate the molarity of OH-, H+ , pH and pOH
a. 0.01M NaOH
[OH-] = 0.01
pOH = 2
[H+] = 1x10-14/.01 = 1x10-12
pH = 12
b. 5M HNO3
[H+] = 5 M
[OH-] = e-14 / 5 = 2x10-15
pOH = 14.7
pH = -.7
c. 0.007M Ba(OH)2
[OH-] = .014 M
[H+] = 1x10-14/ .014 = 7.14x10-13
pOH = 1.85
pH = 12.15
pf3

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Worksheet 5. Aqueous Equilibrium Problems; Simple Equilibria

  1. Identify the acid/base and their conjugate base/acid, and which definition you use to determine(Bronsted, Arrhenius or Lewis): a. HCO3- + H+ ↔ H2CO Base conj acid: Bronsted b. HCO3- ↔ CO 3 2- + H+ Acid conj base : Arrhenius c. CH3NH2 + H2O ↔ CH3NH3+ + OH- Base acid conj acid conj base : Lewis d. C6H5OH + H2O ↔ C6H5O- + H3O+ Acid base conj base conj acid : Lewis, Arrhenius, Bronsted e. H2O + H2O ↔ H3O + + OH- Acid base conj acid conj base
  1. Assuming Kw = 1x10-14^ , calculate the molarity of OH -^ in solutions at 25ºC when the H+^ concentration is: a. 0.2M At 25ºC, Kw = [OH-] [H+] = 1x10 - [OH-] = 1x10-14^ / .2 = 5x10- b. 5 x 10-10^ M [OH-] = 1x10-14^ / 5 x 10-10^ = 2 x 10 - c. 100 M [OH-] = 1x10-14^ / 100 = 1x10-
  2. For each of these strong acid/base solutions, calculate the molarity of OH-, H+^ , pH and pOH a. 0.01M NaOH

[OH-] = 0. pOH = 2 [H+] = 1x10 -14^ /.01 = 1x10- pH = 12

b. 5M HNO (^3)

[H+] = 5 M [OH-] = e-14 / 5 = 2x10- pOH = 14. pH = -.

c. 0.007M Ba(OH) (^2) [OH-] = .014 M [H+] = 1x10 -14^ / .014 = 7.14x10- pOH = 1. pH = 12.

  1. Fill in the blank (all concentrations are at equilibrium): HA ↔ H+^ + A- Acid [HA] [H+^ ] [A-^ ] Ka Chlorous acid HClO 2 0.6 .077 .077 1.0 x10-

Nitrous acid 18.6 .20 4 4.3 x10-

Hydrocyanic acid HCN 28 7 x10-5^ 2 x10-4^ 4.9 x10-

Phosphoric acid H 3 PO 4 0.076 0.3 .00192 7.6 x10-

  1. The pH of a 0.115M solution of chloroacetic acid, ClCH 2 COOH, is measured to be 1.92. Calculate Ka for this monoprotic acid.

pH = -log[H3O+] [H3O+] = 10-pH = 10-1.92 = 0.012 M

HA + H2O → H3O+ + A Initial 0.115 M 0 M 0 M Change -0.012 M +0.012 M +0.012M Equilibrium 0.103 M 0.012 M 0.012 M Ka = [H3O+][A-]/[HA] = (0.012)(0.012)/0.103 = 1.4 x 10 -

  1. Calculate the concentrations of all the species and the pH in 0.10 M hypochlorous acid, HOCl. For HOCl, Ka = 3.5 x 10-^.

HOCl + H2O ↔ H3O+ + OCl- Initial 0.10 M 0 M 0 M Change -x M +xM +xM Equil (0.10 – x)M xM xM

Ka = [H3O+][OCl-]/[HOCl] = (x)(x)/(0.10-x) = 3.5 x 10- x2/0.10 ≈ 3.5 x 10-8 x2 ≈ 3.5 x 10-9 x2 ≈ 5.9 x 10- [H3O+] = xM = 5.9 x 10-5 M [OCl-] = xM = 5.9 x 10-5 M [HOCl] = (0.10 – x) M = (0.10 – 0.000059) M = 0.10 M [OH-] = 1x1014/[H3O+] = 1x1014/5.9 x 10-5 1.7 x 10-10 M pH = -log(5.9 x 10-5) = 4.

  1. One liter of saturated barium sulfate solution contains 0.0025 grams of dissolved BaSO 4. (a) What is the molar solubility of BaSO 4? BaSO4(s) ↔ Ba2+(aq) + SO42-(aq) Ksp = [Ba2+][ SO42-]

0.0025 g BaSO4/1.0 L x 1 mol BaSO4/233 g BaSO4 = 1.1 x 10-5 mol BaSO4/L (dissolved)

(b) Calculate the solubility product constant for BaSO4. BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)