Honors Chemistry



Honors Chemistry

Chapter 19 Notes – Part 2 – Acids, Bases, and Salts

(Student edition)

Chapter 19 Part 2 problem set: 65, 69, 89, 93, 94, 96, 98, 99

19.2 Hydrogen Ions and Acidity

Tap water . Why? Tap water contains many ( )

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 = don’t include

we call this constant

At 25 C0, [H3O+1] = so [OH-1] =

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.

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 =

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:

Acid Neutral Base

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 | | |

To convert pH to [H3O+1]:

Formula: [H3O+1] = Antilog -pH

Calculator: Press the 2nd key, the log key, the (-) key, enter pH, and press the enter key.

Fill in the following table:

|[H3O+1] |pH |Acid or Base |

| |2 | |

| |11 | |

| |5.22 | |

Another formula to know: pH + pOH = 14

|[H3O+1] |[OH-1] |pH |pOH |Acid or Base |

|2.00 x 10-5 | | | | |

| |4.10 x 10-5 | | | |

| | |6.80 | | |

| | | |11.2 | |

NIB - Gram Equivalent Mass

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 =

(

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: ½ mole of H2SO4 is necessary to balance 1 mole of NaOH.

½ mole H2SO4 =

(

Ex3: To make H3PO4 chemically equivalent to NaOH, 1/3 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 |1 |1 |1 |

|Equivalent Weights |36.5 |49.05 |32.7 |

Formula for calculating equivalent weight:

eq. wt. = mw/equivalents

: the # of hydrogen moles or hydroxide moles.

Formula for calculating equivalents:

# equivalents = (moles)(n)

n = # of H or OH in the chemical formula

Ex1a: Calculate the molecular weight of H2CO3: mw =

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?

NIB - Normality In the past, we have used for concentration. M = moles / L

A more useful form of concentration for acid/base reactions is Normality.

N =

Also, in calculating pH, normality is used instead of molarity.

but, Normality is related to Molarity: N =

Ex1: Calculate the Normality and Molarity if 1.80 g of H2C2O4 is dissolved in 150 mL of

solution.

Section 20.11 is not yet finished, but we’ll finish it later….

NIB – problems involving mixing unequal amounts of acid and base

Find the pH of a solution made by mixing 50.0 mL of .10 M HCl with 49.0 mL of .10 M NaOH

Find the pH when mixing 50.0 mL of .10 M sulfuric acid with 50 mL of 1.0 M NaOH

19.2 Indicators

Indicators: dyes where the color depends on the amount of ion present. They are used to show the of solutions.

Indicators are usually made up of . The general formula is…

HIn ( H+ + In-1

Color 1 Color 2

(acid color) (base color)

How do they work?

If the solution is basic solution (lot’s of ):

When an indicator is added, H+ from the indicator reacts with OH- to make . Now, all the H+ on the right side is . Since the equilibrium is disturbed (Le Chatelier), the reaction shifts to the . This makes more . More means more color.

If the solution is acidic (lot’s of ):

When the indicator is added to the solution, the high amount of H+ causes the equilibrium to shift . This makes more . More means more like the color.

Indicators and colors to know: Phenolphthalein, Bromthymol Blue, Methyl Orange,

Litmus Paper

Indicators change over small ranges (Phenolphthalein changes 8.2 - 10.6)

: pH range over which an indicator changes .

19.4 Acid Base Neutralization Acid + Base (

Ex1: Write the molecular, total ionic, and net ionic equations for the neutralization of

HCl with NaOH.

(molecular)

(total ionic)

(net ionic)

Note: getting a perfect match of H3O+1 and OH-1 is next to impossible, so most neutral solutions are slightly acidic or basic.

Ex2: Write the molecular, total ionic, and net ionic equations for the neutralization of

H2SO4 with NaOH.

(molecular)

(total ionic)

(net ionic)

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.

Standard Solution: the solution with a concentration.

A graph of the titration of HCl with NaOH:

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?

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.

now let’s go backwards

Ex6: How many mL of a 1.20 % HCl solution are needed to titrate 25.50 mL of

0.100 M magnesium hydroxide?

19.5 Salts in Solution

Hydrolysis – the reaction of a substance with water

acids – acidic bases – basic

salts – neutral? Sometimes, but they can lead to acidic/basic solutions

hydrolysis of the salt of a strong base and weak acid

NaC2H3O2 + H2O → Na+ + C2H3O2-1

Since NaOH is a strong base, there is no attraction of sodium ion for the hydroxide ion produced from the self ionization of water, but….

C2H3O2-1 is the salt of a weak acid so H+ (from the self ionization of water) is attracted to the acetate ion.

H+ + C2H3O2-1 ↔ HC2H3O2

Now since there is less H+, more H2O will self ionize creating more H+ (which will also get removed) and more OH-1 making the solution basic.

Hydrolysis of the salt of a strong acid and weak base

NH4Cl + H2O → NH4+ + Cl-1

NH4+ + OH-1 ↔ H2O + NH3

so… hydroxide is taken out of solution. The equilibrium will shift when water self ionizes creating more OH- (which will also get removed) and more H+1 making the solution acidic.

Hydrolysis of the salt of a weak base and weak acid

(NH4)2CO3 - son of NH4OH and H2CO3

( could produce a solution that is acidic, basic or neutral)

NH4+ + OH-1 ↔ NH3 + H2O

CO3-2 + H+1 ↔ HCO3-1

Since both of these reactions happen to some degree, it is hard to tell which happens more – therefore this solution can be acidic , basic, or neutral.

Hydrolysis of the salt of a strong acid and strong base

Cl-1 + H+1 does not yield HCl

Na+1 + OH-1 does not yield NaOH

Neither of these reactions happen because nature does not favor the production of strong acids/bases. Therefore, the amount of H+1 and OH-1 ions created from the self ionization of water does not change. A neutral solution results.

What kind of solution would be produced by the hydrolysis of the following salts?

NH4Br K2SO4 CaCrO4 Na2CO3 Fe(C2H3O2)3

acid neutral basic basic can’t tell

4. Buffer solutions – sometimes solutions need to be made “resistant to change” in pH – an example is blood – the pH can vary between 7.3 and 7.5 but…

under 6.9 – acidosis = above 7.7 – alkalosis =

so… blood has buffers to help keep pH relatively constant

An example of a buffer solution: HC2H3O2 and NaC2H3O2

normally HC2H3O2 + H2O ↔ H3O+ + C2H3O2-1

(slightly acidic)

normally NaC2H3O2 + H2O → Na+ + C2H3O2-1

(hydrolysis – salt of a SB/WA – solution is slightly basic)

so…. when HCl is added to the solution, H+ from the HCl combines with the acetate from the salt neutralizing it

so…. when NaOH is added to the solution, OH-1 for the base combine with H+ from the acetic acid neutralizing it

so…. this buffer solution is resistant to change in pH

buffers are made from weak acids and the salt of that weak acid (buffer in the acid range)

or

and weak base and its salt ( buffer in the base range)

-----------------------

pH

Addition of base

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