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GWU / Chemistry / CHEM 1112 / What refers to the number of moles of solute in one liter of a saturat

What refers to the number of moles of solute in one liter of a saturat

What refers to the number of moles of solute in one liter of a saturat

Description

School: George Washington University
Department: Chemistry
Course: General Chemistry II
Professor: J kostal
Term: Fall 2015
Tags:
Cost: 50
Name: Study Guide 3
Description: This study guide covers new material for this exam (chapter 15-17). However, please read the other study guides for previous chapters because this is a cumulative test.
Uploaded: 04/14/2018
5 Pages 84 Views 2 Unlocks
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CHAPTER 15Don't forget about the age old question of What do you call a mechanism of evolution where differential survival and reproduction of organisms allow some to survive and reproduce more effectively than others?

15.1

  • Molar solubility : the % of moles of solute in 1L of a saturated solution
  • Solubility is the # of grams of solute in 1L of a saturated solution
  • Solubility product expression and Ksp

Aga(s) ⇌ Ag+ (aa) + c1- (aa)Don't forget about the age old question of Calculate the ph of a solution containing a weak acid or base and its conjugate base/acid.

        Ksp = [Ag+] [C1-]

  • Example : molar solubility of AgCI in pure H2O

        AgCI + H2O ⇌ Ag+ + CI-Don't forget about the age old question of What are the consequences placed upon deviants by others?

        AgCI(s) ⇌ Ag+ + CI-

        Ksp =  [Ag+] +  CI- = 1.6 x 10-10If you want to learn more check out What is the meaning of vocal qualifiers?

        [Ag+] = [CI-] = x We also discuss several other topics like What is the meaning of centralized water treatment systems?
We also discuss several other topics like What is the meaning of actin in biology?

                        = 1.26 x 10-5

  • How about Pbf? Ksp = 3.6 x 10-8

        PdF2 ⇌ Pb2+ + 2F-

        Ksp = [Pb2+] [F-]2

        = (x)(2x)2 = 4x3

           x = 2.0 x 10-3

[Pb2+] = 2.0 x 10-3 mo1/L

[F-] = 4.0 x 10-3 mo1/L

-Ksp and common ion effect

  • Example : calculate molar solubility of CaF2 in a solution that is .05M in ca(CH3(00)2

Ksp = [Ca2+] [F-]2

        (x)    (2x)2   = 4x3

        4.0 x 10-11 = 4x3

                x = 2.15 x 10-4 mo1/L ← in pure H2O

CaF2 ⇌ Ca2+ + 2F-

I             -         0.05           -

C             -          +x       +2x

E             -          .05+x    2x

        4.0 x 10-11 = (.05 + x) (2x)2 assume x is small

                X = 1.4 x 10-5 mo1/L

  • Factors that affect solubility : ph
  • Molar solubility of Mg(OH)2 (s)

Mg(OH)2 ⇌ Mg2+ + 20H-

Ksp = [Mg2+] [OH-]2

        x        (2x)2     = 4x3

        = 1.7 x 10-4 mo1/L

  • pH of Mg(OH)2

[OH-] = 2[Mg2+]

         = 3.4 x 10-4

         pOH = 3.46

         pH = 10.53

  • Molar solubility of Mg(OH)2 in solution is buffered @ pH = 9

Ksp = [Mg2+] [OH-]2                        pH = 9

1.8 x 10-11 = [Mg] [1.0 x 10-5]2                pOH = 5

[Mg] = .18M                                [OH-] = 10-5 = 1.0 x 10-5

  • In general, the solubility of a compound containing a basic anion increases as solution becomes more acidic
  • Selective precipitation
  • Example 1 (Slide 12)
  1. Baso4 ⇌ Ba2+ + SO42-                                Ksp

Ksp = [Ba2+] [SO42-] = x2                        mo1 so1 = 1.05 x 10-5

        SrSO4 ⇌ Sr2+ + SO42-

                Ksp = [Sr2+] [SrO42-] = x2                        mo1 so1 = 5.65 x 10-4

                3.2 x 10-1 = .01 [SO42-]

Because Barium has a smaller mo1 so1. It precipitates 1st.

b. Ksp = [Ba2+] [SO42-]

        1.10 x 10-10 = [.01] [SO42-]

        1.1 x 10-8 = [SO42-] when BaSO4 will begin to precipitate

Q = [Ba2+] [SO42-]

                ⤷ 0

        Q = 0         adding SO42-, increase Q

        Q < K

c. Ksp = [Sr2+] [SO42-]

        = (.01)[[SO42-]]

        3.2 x 10-5         M = [SO42-]

Baso4

        Ksp = [Ba2+] [SO42-]

               = [Ba2+] [3.2 x 10-5]         x 100

                                        .034%

               = 3.43 x 10-6 = [Ba2+]      ⤴

  • Example 3 (slide 10)

(aF2(s) ⇌ Ca2+ + 2F-

        Q = [Ca2+] [F-]2

[Ca2+] =  = .1moI / L Ca2+

[F-] =  = .04 moI / L F-

        Q = [.1][.04]2

        Q = 1.6 x 10-4

CaF2 ⇌ Ca2+ + 2F-

        Ksp = 4 x 3

        X = 2 x 10-4 moI / L

Q > K ➝ Q is way bigger than K so its supersaturated

Precipitates until Q = k

  • Example 4 (slide 11)

                        Mg (OH)2 ⇌ Mg2+ + 2OH-         6.3 x 10-10

                        Cu (OH)2 ⇌ Cu2+ + 2OH-

                        

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