Equilibria: Acids and Bases

Group Work.7

 

Zen Moment

In silence contemplate the floating blossom in water.

During the first minute, allow your mind to go where it wants.

During the second minute, become aware-think-about what caused the flower to be.

 

Topic: Conquering the Unknown

A. What is the pH of 100 mL of 0.100 M aqueous solution of NaOH ?

NaOH is a strong base and dissociates completely in an aqueous solution

pOH = -log [OH] = -log 0.1 = 1

pH = 14-1 = 13

B. Calculate the pH after each addition below:

(1). 5.0 mL of 1.00 M HCl is added to the 100mL of 0.10 M NaOH above. pH = 12.7

See next page for solutions.

(2). 3.0 mL more of 1.00 M HCl is added (for a total of 8.0 mL of the HCl) to the 100mL of 0.10 M NaOH above. pH = 12.3

 

(3). 1.0 mL more of 1.00 M HCl is added (for a total of 9.0 mL of the HCl) to the 100mL of 0.10 M NaOH above. pH = 12.0

 

(4). 1.0 mL more of 1.00 M HCl is added (for a total of 10.0 mL of the HCl) to the 100mL of 0.10 M NaOH above. pH = 7.0

 

(5). 1.0 mL more of 1.00 M HCl is added (for a total of 11.0 mL of the HCl) to the 100mL of 0.10 M NaOH above. pH = 2.04

 

(6). 4.0 mL more of 1.00 M HCl is added (for a total of 15.0 mL of the HCl) to the 100mL of 0.10 M NaOH above. pH = 1.36

(7). 5.0 mL more of 1.00 M HCl is added (for a total of 20.0 mL of the HCl) to the 100mL of 0.10 M NaOH above. pH = 1.08

(Values in ice tables have units of moles)

OH- + H3O+ => H2O*

B1) Limiting Reagent? H3O+

(i) 0.01 0.005 -

(c) 0.005 0.005 -

(e) 0.005 0 -

Concentration of H3O+: 0.005 mol/(0.100 L + 0.005 L) = 0.0476 M OH-

pH = 14 — (-log 0.0476) = 12.7

B2) Limiting Reagent? H3O+

(i) 0.005 0.003 -

(c) 0.003 0.003 -

(e) 0.002 0 -

Concentration of H3O+: 0.002 mol/(0.105 L + 0.003 L) = 0.0185 M OH-

pH = 14 — (-log 0.0185) = 12.3

B3) Limiting Reagent? H3O+

(i) 0.002 0.001 -

(c) 0.001 0.001 -

(e) 0.001 0 -

Concentration of H3O+: 0.001 mol/(0.108 L + 0.001 L) = 0.00917 M OH-

pH = 14 — (-log 0.00917) = 12.0

B4) Limiting Reagent? None

(i) 0.001 0.001 -

(c) 0.001 0.001 -

(e) 0 0 -

Only source of H3O+ is from the dissociation of neutral water so the pH = 7

B5) Limiting Reagent? OH-

All of the hydroxide ions have been reacted at this point, except the small concentration due to the dissociation of water that can be neglected, so now we calculate the concentration of hydronium ions after each addition, which is just a dilution problem .

Moles H3O+ added: 0.001 L x 1.00 M H3O+ = 0.001 mol H3O+

New Concentration of H3O+: 0.001 mol H3O+/ 0.111 L = 0.00901 M H3O+

pH = -log 0.00901 = 2.04

B6) Limiting Reagent? OH-

Moles H3O+ added: 0.004 L x 1.00 M H3O+ = 0.004 mol H3O+

New Concentration of H3O+: (0.001 + 0.004) mol H3O+/ 0.115 L = 0.0435 M H3O+

pH = -log 0.0435 = 1.36

B7) Limiting Reagent? OH-

Moles H3O+ added: 0.005 L x 1.00 M H3O+ = 0.005 mol H3O+

New Concentration of H3O+: (0.005 + 0.005) mol H3O+/ 0.120 L = 0.0833 M H3O+

pH = -log 0.0833 = 1.08

* (Values in ice tables have units of moles)

C. Plot your results, pH vs volume of HCl added below.

 

D. Mystery Solid Leads to a Crime Solved!!!

You the chemist can make the identification.!!!!

...near the slain body of .... was an old, brown glass bottle covered in dust with a barely legible---actually, it _was_ illegible---label. Inside the bottle was a trace amount of white crystalline solid. You decided not to taste it. The label looked like this:

You the famous forensic chemist knew if you could identify the material you could track and find the murderer and solve the heinous crime. You realize a neutralization reaction of the basic component in the solid by reaction with an acid could give you the molecular weight and then you could identify its formula. So you take the remaining 1.0 g, dissolve it in water and titrate it with 0.500M HCl. Your phnelthalein indicator turned colorless after 23.3 mL HCl was added. What is the mystery solid?

The neutralization (stong acid and strong base) reaction is:

2HCl + _?_(OH)2 => _?_ + 2Cl- + 2H2O

Calculate the moles of HCl:

0.500 M HCl x 0.0233 L = 0.0116 mol HCl at equivalence point

Calculate the moles of the remaining unknown material:

0.0116 mol HCl x (1 mol OH-/1 mol HCl) x (1 mol _?_(OH)2/2 mol OH-)

= 0.00582 mol _?_(OH)2 in remaining sample

Calculate molecular weight of _?_(OH)2:

MW = 0.00582 mol _?_(OH)2/1.0 g = 171.7 g/mol

Calculate molecular weight of _?_:

MW(_?_) + MW(2xOH) = MW(_?_(OH)2) = 171.7 g/mol

MW(_?_) = 137.7 g/mol where MW(2xOH) = 34 g/mol

Using the ever handy periodic table…the unknown compound is Ba(OH)2