Honors Chemistry



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 Base |

|1 |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 Base |

|1 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 Base |

| |2 | |

| |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 Base |

| | | | | |

|2.00 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 |H3PO4 |

|Moles 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

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pH

Addition of base

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