Chemistry 115

Quiz 1

 

April 12, 2001

 

Name _________________________________
  1. An aqueous solution contains 15.0 g of sugar, C12H22O11 (MM = 342.3), in 120 mL. The density of this solution is 1.047 g/mL. Calculate mol = 15.0 g/342.3 g/mol = 0.04382 mol (6)

Mass = 120 mL x 1.047 g/mL = 125.64 g

  1. The weight per cent of the sugar

    2. Wt% = mass sugar/mass solution x 100% = 15.0 g/125.64 g x 100% = 11.939%

    Wt % = |_11.9%__

  2. The molarity of the sugar M = mol solute/L solution

    3. = .04382 mol / 0.120 L = 0.36518 M

    M = |_0.365 M___

  3. The molality of the sugar m = moles solute / kg solvent

mass solvent = 125.64 g - 15.0 g = 110.64 g

m = 0.04382 mol / 0.11064 kg = 0.3961 m m = |_0.396 m___

 

  1. Calculate the pH of the final solution after 10.0 mL of 1.0 M NaOH is added to 100. mL of a buffer solution that is 0.20 M propionic acid and 0.15 M sodium propionate? Ka = 1.3 x 10-5 =[H+][P-]/[HP] (4)

    2. 10.0mL x 1.0M = 10mmol OH-

    100mL x 0.20M = 20mmol HP

    100mL x 0.15M = 15mmol P-

    HP +

    OH- =

    P- +

    H2O

    20mmol - 10mmol = 10 mmol

    10mmol -10mmol =0 mmol

    15mmol +10mmol = 25 mmol

     

    Ka = 1.3 x 10-5 = [H+][P-] / [HP] = [x][25/110 + x] / [10/110 -x] ~ [x][25] / [10]

    [H+] = x = 1.3 x 10-5 x 10 / 25 = 5.2 x 10-6 M; pH = 5.283

    pH = |_5.28_____

  2. At 25oC, 1.7 x 10-5 mol of Cd(OH)2 dissolves in one liter of water. Calculate the solubility product for Cd(OH)2. (3)

    3. Cd(OH)2(s) =

    Cd2+ +

    2 OH-

    (-1.7 x 10-5)

    1.7 x 10-5

    2(1.7 x 10-5)

    Ksp = [Cd2+][OH-]2 = [1.7 x 10-5][2(1.7 x 10-5)]2 = 4 x (1.7 x 10-5)3 = 1.9652 x 10-14

    Sol.Prod. = |_2.0 x 10-14 _

     

  3. Mercury(I) chloride, Hg2Cl2 is an unusual salt in that it dissolves to form Hg22+ and 2 Cl-; Ksp= 1.3x10-18

Calculate Hg2(s) = Hg22+ + 2 Cl- (7)

  1. the molar solubility of Hg2Cl2 in distilled water

    2. Ksp = 1.3 x 10-18 = [Hg22+][Cl-]2 = [x][2x]2 = 4 x3

    X = (1.3 x 10-18 / 4)1/3 = (3.25 x 10-19)1/3 = 6.875 x 10-7

    Sol. = |_6.9 x 10-7 M

  2. the molar solubility in a 0.10 M FeCl3 solution. à Fe3+ + 3 Cl-; [Cl-] = 3 x 0.10 = 0.30 M Cl-

Ksp = 1.3 x 10-18 = [Hg22+][Cl-] = [x][2x + 0.30]2 ~ [x][0.30]2

X ~ 1.3 x 10-18 / [0.30]2 ~ 6.875 x 10-7 M

Sol = |_1.4 x 10-17 M_

 

Bonus: Calculate the molar solubility of Ag4Fe(CN)6 (= 4 Ag+ + Fe(CN)64-) given its solubility product is 1.6 x 10-41. (2) Happy Easter!

Ksp = 1.6 x 10-41 = [Ag+]4[Fe(CN)64-] = [4x]4[x] = 256 x5

X = (1.6 x 10-41 / 256)1/5 = (6.25 x 10-44 )1/5 = 2.2865 x 10-9

Sol.. = |_2.3 x 10-9 M