October 22, 2003

Answer questions 1-9 and any three of the four problems (10-13).

You must show your work to receive full credit for your answer.

I. Short acid / base / solubility calculations 25 points (5 pts each)

1. (5 pts) An aqueous solution has [H₃O⁺] = 2.50 x 10^-5 Molar

what is the pH ?

• pH = -log(2.50x10-3) = +

what is the pH ?

• pOH = - log(3.50 x 10^-5) =
• pH = 14.00- pH = 14.00 - =
• aternative way
• [H+] = 1.0x10-14/3.50 x 10^-5 =
• pH = -log( ) =

a. what is pKa for acetic acid ?

• pKa = -log (Ka)
• pKa= - log(1.5 x 10^-5) =

• Ka * Kb = Kw = 1.00 x10-14
• Kb = 1.00 x 10-14/1.5 x 10^-5

• as an acid it produces H+ or H3O+ in water
• as a strong acid, it is 100% dissociated in water

II. Concepts (20 pts)

• a proton donor

the HCN acts as a __weak acid__.

CN- is said to be the _ conjugate_ __ base__ of HCN

H₃O⁺ is an acid

H₂O is acting as a __base (conjugate base of water)_ in this reaction

0.10 mole of HCl and

0.12 moles of NaOH ... we can expect to end up with approximately

• the strong base reacts first with the stronger acid (the HCl)
• no HCL remains, but we still have 0.02 moles of OH-
• this base reacts with acid (only HAc is left to react)
• all the base is consumed
• we still have 0.080 moles of HAC
• we formed 0.020 moles of acetate ion
• also have 0.10 moles of Na+ and Cl-

______ moles of OH-

______ moles of acetate ion

______ moles of Cl-

• without a massive change in infrastructure the H2 is produced by chemical reaction, fueled by burning coal or by electrical decomposition of water. The electricity comes mainly from coal and oil fired generators. Before the H2 gets to the tank, large quanitites of CO2 have been produced and sent into the atmosphere.
• there are alternative energy sources-- nuclear / wind / hydroelectric produced electricity. However, there's not enough available now to make a dent into the auto fuel program.

There are 4 problems (10-11-12-13) Answer ANY THREE of them

• Ka=1.5 x 10-15 for acetic acid (page 1)
• HA --> H+ + A-
• Ka = [H+]*[A-] / [HA]
• let x= [H+] formed
  • then [A-] = x also
  • [HA] = 0.15 - x
• Ka= x² / (0.15-x)
• simpler if we can neglect x in the term 0.015-x
  • Ka = x² / 0.15
  • x=[H+] = (1.5x10^-15* 0.15)^1/2
  • x=
    • that assumption looks awfully good here
    • of course we could have done the quadratic form (smae final answer)
  • pH = - log( ) =

very similar, but this is a base in water

the base is acetate ion

Kb = (from problem 4 on first page)

• A- --> OH- + HA
• Kb = [OH-]*[HA] / [A-]
• let x= [OH-] formed
  • then [HA] = x also
  • [A-] = 0.15 - x
• Kb= x² / (0.15-x)
• simpler if we can neglect x in the term 0.015-x
  • Kb = x² / 0.15
  • x=[OH-] = (1.xxx10^-xx* 0.15)^1/2
  • x=
    • that assumption looks awfully good here
    • of course we could have done the quadratic form (smae final answer)
  • pOH = - log( ) =
  • pH = 14.00 - xxx = xxx

a. what is Ka for this acid

b. what is the degree of dissociation of the acid

(Na₂SO₄) solution? Ksp= 1.2 x 10^-5 for Silver Sulfate