Weak acid, weak base

Acids = proton donors.
Weak acids - those that are not strong!
Weak acids do not ionize 100%.
Weak acids set up equilibrium with water to produce the acid's anion and hydronium ion.
If the general representation of a weak acid is HA, then in llbm:
HA(aq) + H2O <=>  H3O+(aq)  +  A-(aq)

Generally two question types are asked
(1) Determine the Ka value for a particular weak acid
   -given llbm information about [H3O+] via pH and/or %ionization information
   -where pH can be used to determine [H3O+] @ llbm
   -and where %ionization = [H3O+]llbm/[HA]  x 100%
(2) Determine pH (or % ionization) given specific acid(useKa chart to determine Ka value)

Complete mathematical solution would include:
-llbm equation
-Ka expression
-ICE chart
-solve for whatever it is that is being asked
     -looking for pH or %ionizn?
     -solve for [H3O+]llbm, then calculate to get requested info
     -looking for Ka, use pH or %ionzn to determine [H3O+]

Weak bases - similar set up as with weak acids
Difference, weak bases involve [OH-], which leads to pOH

If the geneal formula for a weak base is 'Base' and bases are proton acceptors
Generally: Base(aq)  +  H2O  <=>  BaseH+(aq)   +   OH-(aq)
Again given two basic question types
(1) Determine pH of a stated basic solution.
(2) Determine Kb of a stated base.

Complete mathematical solution would include:
    -llbm equation
    -Kb expression
    -ICE chart
    -solve for whatever it is that is being asked
         -looking for pH?
             -solve for [OH-]llbm , then determine pOH, then pH
              @25C pH  + pOH = 14.00
         -looking for Kb use given pH (or pOH) to get to [OH-]llbm

Homework:
p568(8) p570(10)  p574(12,13)  p579(1,3,13)
Next up?
Using the Ka/Kb chart to calculate K values of conjugate ions
KaKb = Kw

Cute joke coming, ready? Why did the chicken cross the playground? [Ans: To get to the other slide.]

Acid- Base: pH/pOH Calculations

Strong Acids
If it is not a stong acid, it is a weak acid.
Know the strong acids (monoprotic and diprotice)
Strong acids ionize virtually 100%

HCl(aq)  + H2O => H3O+(aq)  +  Cl-(aq)

pH = -log[H3O+]; 10^-pH = [H3O+];


Question: Determine the pH of a 0.150M HCl solution
(since strong acid ionizes 100%, [HCl] = [H3O+]=0.150M

pH = -log(0.150M) = 0.824
Notice: 3 sgfigs in concentration 0.150M; 3 decimal places
 in pH value 0.824

Determine the concentration of a monoprotic acid solution if the pH
of the solution is 4.35
10^-4.35 =4.5x10^-5  M
Notice: 2 decimal places n pH value, 2 sigfigs in concentration answer

Only the decimal places of pH values are significant.


In a similar way that pH/[H3O+] connect, so can
pOH and [OH-]

pOH = -log[OH-] and 10^-pOH = [OH-]

Connecting pH and pOH?

@25C:   pH + pOH = 14.00

Question: Determine the pH of a 0.250M NaOH solution
Strong bases dissociate virtually 100%
NaOH is a strong base, dissociates 100%
NaOH(aq) => Na+(aq)  +   OH-(aq)
So, [NaOH] = [OH-] = 0.250M

When working with a base, one cannot determine pH directly from the given base concentration . Determine pOH, then determine pH

pOH = -log(0.250M) = 0.602
pH = 14.00-0.602 = 13.398

there is another way to carry out this calculation
- saving that for another time

A few basic jokes to end this post: What happened when the lion ate the comedian? Ans: It felt funny. Where do sheep go to get a haricut? Ans: To the Baa Baa shop. They are simple, basic jokes. Get it? This post dealt with bases, so it ended with some basic jokes!

llbm Review

-complete the llbm & solubility review handout
-check moodle for introduction to acid/base equilibrum handouts
-check moodle for upcoming quiz/test dates

Will a precipitate occur?

Will 35.0mL of 0.0100M NaCl form a precipitate when combined
with 65.0 mL of 0.0150M AgNO3?

• write the double displacement to identify the slightly soluble salt (the precipitate)
• silver chloride in the example
• determine the concentration of the ions of the slightly soluble salt using c1V1=c2V2
• Cl and Ag in this example
• for Cl: (0.0350L)(0.0100M)=c2(0.100L*);
• * 35mL + 65mL = 100mL = 0.100L
• Write the llbm equation for the slightly soluble salt then the Ksp expression
• Use the calculated concentrations, sub into Ksp to determine a "trial K"
• compare "trial K" to real K (from Ksp chart)
• if trial K less than real K = no precipitate (not enough of the ions present to cause a ppt to form)
• if trial K greater than real K = ppt (too great of an amount of the ions in the solution - excess ppts out)

Remember if you are not part of the solution, you are part of the precipitate.

Lab Follow-Up

The lab involved neutralizing 25.00mL of an unknown concentration of a calcium hydroxide solution with a determined volume (via the buret) of 0.100M hydrochloric acid.

The solution is already at equilibrium when the titrating begins.
With the titration data it is possible to:

• determine molar solubility of the calcium hydroxide in pure water
• determine Ksp value for calcium hydroxide
• determine molar solubility of calcium hydroxide in a calcium chloride solution

There was excess calcium hydroxide in the solution. the mass of calcium hydroxide is not relevant to the lab follow-up calculation. Furthermore, the calcium chloride was added to upset equilibrium. However, the mass of added calcium chloride is not relevant to the lab follow-up calculations.

The titration data is all that is needed to complete the calculations.


Next up - Acid-Base llbm. Two introductory sheets are on moodle. Notice the moodle calendar for upcoming quiz dates.


Talent alone won't make you a success. Neither will being in the right place at the right time, unless you are ready. The most important question is: 'Are your ready?' Johnny Carson Are you ready for acid-base solution equilibrium?

Common Ion Effect

Homework:
p492 (9)
P492 (10,11,12)

Determine the molar solubility of Fe(OH)3 in a 0.300M Fe2(SO4)3 solution.

      Fe(OH)3(s)   <=>     Fe3+ (aq)      +       3 OH-(aq)

I       n/a                           0.600M*                    0
C     n/a                                 +x                       +3x
E     n/a                            0.600 + x                  +3x
                                  ~0.600 (small K)

initial Fe3+ = 0.600M - from the 0.300M Fe2(SO4)3,
where there are twice as many Fe as Fe2(SO4)3 units

Ksp = [Fe3+][OH-]3
2.6x10-39 = (0.600)(3x)3
2.6x10-39 = 0.600(27x3)
5.4x10-14 = x 
molar solubility of Fe(OH)3 in the 0.300M Fe2(SO4)3 solution
 is 5.4x10-14 mol/L

Molar Solubility to Ksp

Molar solubility is measured at llbm.

question: Determine the  Ksp of salt AB2 if the molar solubility is
 0.00015 mol/L

1.Write the llbm dissociation rxn.
2. Write the Ksp expression.
3. Set up the ICE chart.
4. Use molar solubility to determine [ion]llbm
5.Sub into Ksp expression.

    AB2 (s)     <=>    A+(aq)   +   2B-(aq)                        Ksp=[A+][B-]2
I    n/a                      0              0
C  n/a
E   n/a                0.00015*        2(0.00015)
* for every mole of A+ that is present at llbm, one mole of AB2 dissolved

Ksp = (0.00015)(0.00030)2
Ksp = 1.4x10-11 for AB2

Sometimes the molar solubility info can be 'hidden':
for example, a saturated solution contains 0.24g of salt per 500mL.
[Use molar mass, convert the given info to mol/L]

Homework:
p486(4)
p493(7,8,9,12)
 ...not all are molar solblty to Ksp, some are Ksp to molar solblty
What would you get if image 1 and  image 2 were combined.

[Ans: a molar solution.]