Accelerated Chemistry Chapter 15 Notes
(Student’s edition)
Chapter 15 bookwork: page 523: 6 -13, 17, 20, 25, 26
Self Ionization of Water
Tap water .
Why? Tap water contains many (F-1 and Cl-1)
Distilled water appears to conduct electricity, but it does – just a little, tiny bit.
H2O + H2O
The reaction happens 0.0000002% in the direction and 99.9999998%
in the direction.
The equilibrium constant for the self ionization of water:
Keq = [H3O+1][OH-1]
don’t include
we call this constant
At 25 C0, [H3O+1] = 1 x 10-7
so [OH-1] = 1 x 10-7
so Kw =
Example: If the [H3O+1] is 1 x 10-3M, then what is the [OH-1]?
The solution is because the hydronium ion concentration is than
the hydroxide concentration.
Page 1
Fill in the table below:
Beaker # / [H3O+1] / [OH-1] / Acid or Base1 / 1 x 10-5
2 / 1 x 10-2
3 / 2 x 10-4
4 / 4.16 x 10-6
Of course, all these numbers are confusing so…
The pH of a Solution
pH = - log [H3O+1]
pH stands for .
logs are functions of . For example, the log of 1000 is (like ) and the log of .01 is (like ).
Acid – Base scale:
0------7------14
really it is from –2 to 16 since no acids/bases ever get more than 100 M solutions
To convert [H3O+1] to pH:
Formula:
pH = - log [H3O+1]
Calculator:
Press the (-) key, the log key, enter the [H3O+1], and press the enter key.
Fill in the following table:
[H3O+1] / pH / Acid or Base1 x10 –1
1 x 10 –9
3.00 x 10 –4
Page 2
To convert pH to [H3O+1]:
Formula:
[H3O+1] = Antilog -pH
Calculator:
Press the 2nd key, the log key, the (-) key, enter the pH, and press the enter key.
Fill in the following table:
[H3O+1] / pH / Acid or Base2
11
5.22
All formulas to know:
Find pH / Find pOH / Find [H3O+1] / Find [OH-1]pH = - log [H3O+1] / pOH = - log [OH-1] / [H3O+1] = Antilog -pH / [OH-1] = Antilog -pOH
pH = 14 – pOH / pOH = 14 – pH / [H3O+1] = 1 x10 –14 / [OH-1] / [OH-1] = 1 x10 –14 / [H3O +1]
Examples:
[H3O+1] / [OH-1] / pH / pOH / Acid or Base2.00 x 10-5
4.10 x 10-5
6.80
11.2
Page 3
Concentration units for Acids and Bases
Chemical Equivalents: quantities of solutes that have .
Ex1:HCl + NaOH NaCl + H2O
To achieve a balance, 1 mole H+ needs to cancel out 1 mole OH-1.
So, for the above reaction:
1 mole of HCl is necessary to balance 1 mole of NaOH.
1 mole HCl = 1 mole NaOH
Ex2:H2SO4 + 2 NaOH Na2SO4 + 2 H2O
To achieve a balance, 1 mole H+ needs to cancel out 1 mole OH-1.
So, for the above reaction:
___ mole of H2SO4 is necessary to balance 1 mole of NaOH.
½ mole H2SO4 = 1 mole NaOH
Ex3: To make H3PO4 chemically equivalent to NaOH, ___ mole of H3PO4
balances 1 mole NaOH
Equivalent Weight: the # of grams of acid or base that will provide of protons or
hydroxide ions.
HCl / H2SO4 / H3PO4Moles of Acid / 1 / ½ / 1/3
Moles of Hydrogen
Equivalent Weights
Formula for calculating equivalent weight:
eq. wt. = mw/equivalents
Page 4
Equivalents: the # of moles or moles.
Formula for calculating equivalents:
# equivalents = (moles)(n) where n = # of H or OH in the chemical formula
Ex1a: Calculate the molecular weight of H2CO3:
Ex1b: Calculate the # of moles in 9.30 g of H2CO3:
Ex1c: Calculate the equivalent weight of H2CO3:
Ex1d: How many equivalents in 9.30 g of H2CO3?
Ex1e: How many grams of H2CO3 would equal .290 equivalents?
Normality
In the past, we have used for concentration. M =
A more useful form of concentration for acid/base reactions is Normality.
N = # eq / L# equivalents = (moles)(n)
n = # of H or OH in the chemical formula
When calculating pH, normality is used over molarity but…
Normality is related to Molarity:
N =
Page 5
Ex1: Calculate the Normality and Molarity if 1.80 g of H2C2O4 is dissolved in 150 mL of
solution.
Continued . . .
NIB – problems involving mixing unequal amounts of acid and base
Ex1: Find the pH of a solution made by mixing 50.0 mL of .100 M HCl with 49.0 mL of
.100 M NaOH.
Ex2: Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of
1.00 M NaOH
Page 6
Acid Base Titration
The object of titration is to get the amount of to equal the amount of .
Titration: the controlled addition of a solution of concentration to a solution of
concentration.
Indicators: dyes where the color depends on the amount of ion present. They
are used to show the of solutions.
Indicators: Phenolphthalein, Bromthymol Blue, Methyl Orange, Litmus Paper
Indicators change over small ranges (Phenolphthalein changes 8.2 - 10.6)
Transition interval: pH range over which an indicator changes .
Standard Solution: the solution with a concentration.
A graph of the titration of HCl with NaOH:
Page 7
Titration Examples:
Ex1: A 15.5 mL sample of .215 M KOH requires 21.2 mL of acetic acid to
titrate to the end point. What is the Molarity of the acid?
That was the difficult way. The easy way …
Normality and Titration:NaVa =NbVb
Ex2: A 15.5 mL sample of .215 M KOH requires 21.2 mL of acetic acid to
titrate to the end point. What is the Molarity of the acid?
Ex3: If 15.7 mL of sulfuric acid is titrated to the end point by 17.4 mL of .0150 M
NaOH, what is the Molarity of the acid?
Page 8
Percent problems:
Ex4: If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid,
calculate the % acetic acid in solution.
That was the difficult way. The easy way …
Ex5: If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid,
calculate the % acetic acid in solution.
Use the titration formula to find the normality of the acid:
NaVa =NbVb
Then use:
(N)(eq wt)/10 = %
Page 9