From WolfWikis

Exercises

Exercise 1

Assuming full dissociation and complete solubility, what is [Cl^-], [La³⁺] and [Y³⁺] in the following aqueous solutions:

1. 0.1 mmol of LaCl₃ in 1 l

   LaCl₃ ==> La³⁺ + 3Cl^-

   [Cl^-]= 3*0.1/1 [mmol]/[l] = 0.3 mM

   [La³⁺]= 0.1/1 [mmol]/[l] = 0.1 mM

   [Y³⁺]= 0

2. 0.05 mmol of LaCl₃ and 0.05 mmol of YCl₃ in 1 l

   LaCl₃ ==> La³⁺ + 3Cl^-

   YCl₃ ==> Y³⁺ + 3Cl^-

   [Cl^-]= 3*0.05+3*0.05 mmol/l = 0.3 mM

   [La³⁺]= 0.05 mmol/1l = 0.05 mM

   [Y³⁺]= 0.05 mmol/1l - 0.05 mM

3. 0.05 mmol of LaCl₃ and 0.02 mmol of YCl₃ in 2 l

   LaCl₃ ==> La³⁺ + 3Cl^-

   YCl₃ ==> Y³⁺ + 3Cl^-

   [Cl^-]= [3*0.05+3*0.03]/2 mmol/l = 0.12 mM

   [La³⁺]= 0.05/2 = 0.025 mM

   [Y³⁺]= 0.02/2 = 0.01 mM

4. 0.05 mmol of LaCl₃ and 0.05 mmol of YCl₃ and 0.05 mmol of BaCl₂ in 1 l

   LaCl₃ ==> La³⁺ + 3Cl^-

   YCl₃ ==> Y³⁺ + 3Cl^-

   BaCl₂ ==> Ba²⁺ + 2Cl^-

   [Cl^-]= [0.05*3+0.05*3+0.05*2]/1 mmol/l = 0.4 mM

   [La³⁺]= .05 mM

   [Y³⁺]= .05 mM

5. 0.01 mmol of LaCl₃ and 0.05 mmol of YCl₃ and 0.05 mmol of La(NO₃)₃ in .5 l

   LaCl₃ ==> La³⁺ + 3Cl^-

   YCl₃ ==> Y³⁺ + 3Cl^-

   LaNO₃ ==> La³⁺ + 3NO₃^-

   [Cl^-]= [3*0.01+3*0.05]/0.5 mmol/l= 0.36 mM

   [La³⁺]= [0.01+0.05]/0.5 mmol/l = 0.12 mM

   [Y³⁺]= [0.05]/[0.5] = 0.1 mM

Exercise 2

Assuming full dissociation and complete solubility, what is [SO₄^2-], [Al³⁺] and [K⁺] in the following aqueous solutions:

1. 0.05 mol of Al metal and 0.02 mol of KCl dissolved in 1 l of .2M sulfuric acid

   [SO₄^2-] = 0.2M

   [Al³⁺] = =0.05 M

   [K⁺]= 0.02 M

2. 0.05 mol of Al₂(SO₄)₃ dissolved in 1 l of .1M nitric acid

   [SO₄^2-] = 0.15M

   [Al³⁺] = =0.10 M

   [K⁺]= 0 M

3. 0.05 mol of Al₂(SO₄)₃ and 0.01 mol of K₂SO₄ in 500cm³ of aqueous solution.

   [SO₄^2-] = [3*0.05+0.01]/.5= 0.32 M

   [Al³⁺] = [0.05*2]/.5= 0.2M

   [K⁺]= [0.01*2]/0.5=0.04 M

Exercise 3

Assuming full dissociation and solubility and neglecting the water equilibrium, what is [H⁺] and pH of the following aqueous systems:

1. A solution of .05M nitric acid

   [H⁺]= .05M

   pH= 1.3

2. A solution of .2M sulfuric acid

   [H⁺]= .4M (diprotic!)

   pH=0.4

3. 200ml solution of .05M nitric acid added to 200ml solution of .2M sulfuric acid

   [H⁺]= [0.05+2*0.2]/2 = 0.225M (The volume increases twofold for both solutions)

   pH= 0.6

4. .2 mol of HCl gas dissolved to form 500 ml of solution

   (monoprotic) [H⁺]=(0.2)/0.5= 0.4M

   pH=0.4

Exercise 4

Assuming full dissociation and solubility and neglecting the water equilibrium, what is [OH^-] of the following aqueous systems:

1. 60 mmol of NaOH, Ba(OH)₂ and KOH each, together in 333 ml solution

   [OH^-] = [60+3*60+60]/.333 = 0.72M

2. 622 mg of anhydrous NaOH and 31 mg of anhydrous Ba(OH)₂ in 540 ml of aqueous solution

   [OH^-] = [622/(23+16+1)+2*31/(137.3+2*17)]/0.54 = 29mM

3. 50 gram of a 50% (by weight) solution of NaOH diluted to 1 liter.

   [OH^-] = [25/(23+16+1)]/1 = 0.625M (In 50g of the stock solution there is 25g of NaOH)

Exercise 5

Assuming full dissociation and solubility and neglecting the water equilibrium, what is [H⁺] of the following aqueous systems after partial neutralization:

1. 500 ml of .1M sulfuric acid to which 30 mmol anhydrous NaOH is added.

   0.1M sulfuric has [H⁺]= 0.2M (diprotic)

   0.5 l of the stuff contains .5*0.2 [l]*[mol/l]= 0.1 mol H⁺

   30 mmol anhydrous NaOH produces 0.03 mol OH^-.

   .03 mol OH^- reacts with .03 mol H⁺ to form water.

   Thus 0.10-0.03= 0.07 mol H⁺ is left in 500 ml.

   final [H⁺] = 0.07/0.5 = 0.14M

2. 500 ml of .1M nitric acid to which 10 mmol anhydrous Ba(OH)₂ is added.

   0.1M nitric has [H⁺]= 0.1M (monoprotic)

   0.5 l of this solution contains .5*0.1 [l]*[mol/l]= 0.05 mol H⁺

   .01 mol Ba(OH)₂ gives .02 OH^-, this reacts with .02H⁺ to form water

   Thus 0.03 mol H⁺ is left in 500 ml

   final [H⁺] = 0.03/0.5 = 0.06M

3. a mixture of 250 ml of .1M nitric acid and 250 ml of .2M sulfuric to which 1 g of anhydrous KOH is added.

   250 ml of .1M nitric contains .025 mol of H⁺

   250 ml of .2M nitric contains .1 mol of H⁺

   total 0.125 mol H⁺ in 500 ml

   1 g KOH produces 0.0178 mol H⁺

   after reaction 0.107 mol H⁺ in 500 ml => [H⁺] 0.2M

Exercise 6

1. 5 g of anhydrous NaOH is exposed to an excess of dry HCl gas. What is the mass of water formed and the [Na⁺]. You may neglect the formation of water vapor, the solubility of HCl in water. Also take the volume of the final solution to be the volume of the water formed and assume complete solubility of the NaCl formed.

   when in doubt...

   NaOH+ HCl ==> NaCl + H₂O

   as both reactants are dry the only water around is what the reaction produces!

   5 g of NaOH is 5 /(23+16+1) [g]/[g/mol]= 0.125 mol

   As HCl is in excess, NaOH is limiting, so the amount of water formed is also .125 mol.

   This corresponds to .125*(18) [mol][g/mol]= 2.25 g of water.

   as the density of water is 1 g/ml this gives a volume of 2.25ml

   the amount of NaCl is also .125g

   if everything dissolves (without changing the volume...) this would give .125/0.00225 = 55.5M...

2. Are the above assumptions realistic?

   No!!! It is not possible to dissolve that much salt in water. There is a solubility limit. Besides, the assumption that the sodium and the chlorine ions would not add to the volume is absurd if there are this many of them.