Honors Chemistry



Honors Chemistry

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

(Student’s 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(l) + H2O(l) (

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 Kw

At 25 C0, [H3O+1] = 1 x 10-7

so [OH-1] = 1 x 10-7

so Kw = 1 x 10-14

Example: If the [H3O+1] is 1 x 10-3M, then what is the [OH-1]?

Kw = [H3O+1][OH-1]

The solution is because the hydronium ion concentration is than

the hydroxide concentration.

Fill in the table below using: 1 x 10-14 = [H3O+1][OH-1]

Note: The concentration values are Normality values (more on this later).

|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 =

Note: The concentration values are Normality values (more on this later).

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(pH 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 =

Note: The concentration values are Normality values (more on this later).

Calculator:

Press the (-) key, the log key, enter the [H3O+1], and press the enter key.

Fill in the following table:

Note: The concentration values are Normality values (more on this later).

|[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] =

Note: The concentration values are Normality values (more on this later).

Calculator:

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

Fill in the following table:

Note: The concentration values are Normality values (more on this later).

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

| |2 | |

| |11 | |

| |5.22 | |

All formulas to know:

Note: The concentration values are Normality values (more on this later).

|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:

Note: The concentration values are Normality values (more on this later).

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

| | | | |(look at the pH) |

| | | | | |

|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___ mole OH-1.

So, for the above reaction:

1 mole of HCl is necessary to balance____ mole of NaOH.

1 mole HCl = _ mole NaOH

(

Ex2: H2SO4 + 2 NaOH ( Na2SO4 + 2 H2O

To achieve a balance, 1 mole H+ needs to cancel out __ mole OH-1.

So, for the above reaction:

__ mole of H2SO4 is necessary to balance 1 mole of NaOH.

__ 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 mole 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. =

Equivalents: the # of moles or moles.

Formula for calculating equivalents:

# equivalents =

n = # of or in the chemical formula

Ex1a: Calculate the molecular weight of H2CO3:

H =

C =

O =

mw =

Ex1b: Calculate the # of moles in 9.30 g of H2CO3:

Ex1c: Calculate the equivalent weight of H2CO3:

eq. wt. = mw/eq

Ex1d: How many equivalents in 9.30 g of H2CO3?

1st convert grams to moles:

Next, convert moles to equivalents:

# equivalents = (moles)(n)

n = # of H or OH in the chemical formula

Ex1e: How many grams of H2CO3 would equal .290 equivalents?

1st convert to moles:

# equivalents = (moles)(n)

n = # of H or OH in the chemical formula

Next, convert moles to grams:

NIB - Normality

In the past, we have used for concentration.

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

N =

# equivalents =

n =

Also, in calculating pH, normality is used over 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.

1st convert grams to moles:

Next, calculate the molarity:

Now, calculate the normality:

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.

Find the moles of HCl and moles of NaOH:

Find the equivalents of H+ in HCl and the equivalents of OH- in NaOH:

eq = (moles)(n)

Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+

and equivalents of OH- from each other (absolute value):

Calculate the normality of the resulting solution:

Calculate the pH:

Ex2: Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of

1.00 M NaOH

Find the moles of H2SO4 and moles of NaOH:

Find the equivalents of H+ in H2SO4 and the equivalents of OH- in NaOH:

eq = (moles)(n)

Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+

and equivalents of OH- from each other (absolute value):

Calculate the normality of the resulting solution:

Calculate the pH:

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)

Transition interval: pH range over which an indicator changes .

Choice of Indicators

There are three main types of titrations:

SA – SB SA-WB WA –WB

Type 1: indicator changes color, solution becomes neutral at the end point, and 1 drop

changes the pH significantly

Type 2: indicator changes color in the acid region – use methyl orange (3.2 to 4.4)

Type 3: indicator changes color in the base region – use phenolphthalein (8.2 to 10.6)

19.4 Acid Base Neutralization

Acid + Base ( Salt + Water

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

HCl with NaOH.

(molecular) HCl(aq) + NaOH(aq) ( NaCl(aq) + H2O(l)

+H2O +H2O

(total ionic) H3O+1(aq) + Cl-1(aq) + Na+1(aq) + OH-1(aq) ( Na+1(aq) + Cl-1(aq) + 2H2O(l)

(net ionic) H3O+1(aq) + OH-1(aq) ( 2H2O(l)

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:

High amount of OH-1

End point

H = OH

High amount of H+1

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?

Calculate the moles of KOH:

Use “stoich” to convert moles of KOH to moles of HC2H3O2:

.

Calculate 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?

Convert molarity to normality:

N = (M)(n)

n = # of H or OH in the chemical formula

Use the titration formula to find the normality of the acid:

NaVa =NbVb

Convert normality to molarity:

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.

Use the titration formula to find the normality of the acid:

NaVa =NbVb

Multiply the normality by the equivalent weight of the acid:

Convert the liters to grams:

Calculate the % by mass:

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

Use:

(N)(eq wt)/10 = %

now let’s go backwards

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

.100 M magnesium hydroxide?

First,

Option #1: Change the % to molarity:

multiply by 10 to get the amount of solute in 1000 grams:

Convert the grams of solute to moles:

Option #2: (N)(eq wt) / 10 = %

Now,

Convert the molarity to normality:

Finally,

Use the titration formula to find the mL of the acid:

1st convert the molarity to normality:

Now, use NaVa =NbVb

19.5 Salts in Solution

Hydrolysis: the reaction of a substance with .

Acids make solutions.

Bases make solutions.

Salts sometimes make solutions. However, they sometimes lead to

solutions.

Ex1: Hydrolysis of the salt of a strong base and weak acid:

NaC2H3O2(s) + H2O(l) ( Na+1(aq) + C2H3O2-1(aq)

The sodium in sodium acetate is from the strong base .

Since NaOH is a strong base, there is no attraction of sodium ion for any

of the present. Hydroxide is present due to the self

ionization of :

H2O(l) + H2O(l) (

but….

The acetate in sodium acetate is from weak .

So, H+1 (from the self ionization of water) is attracted to the ion:

H+1(aq) + C2H3O2-1(aq) ( HC2H3O2(aq)

Now, since there is less , the self-ionization of water reaction will shift

to the . H2O will self ionize creating more (which will also

get ) and more . The increase in makes the

solution .

Ex2: Hydrolysis of the salt of a and :

NH4Cl(s) + H2O(l) ( NH4+1(aq) + Cl-1(aq)

The chlorine in ammonium chloride is from the strong acid .

Since is a strong acid, there is no attraction of chloride ion for any

of the present. Hydrogen ions are present due to the self

ionization of :

H2O(l) + H2O(l) (

but….

The ammonium in ammonium chloride is from weak .

So, OH-1 (from the self ionization of water) is attracted to the ion:

NH4+1(aq) + OH-1(aq) ( NH4OH(aq)

Now, since there is less , the self-ionization of water reaction will

shift to the . H2O will self ionize creating more (which will also get ) and more . The increase in ion makes the solution .

Ex3: Hydrolysis of the salt of a weak base and weak acid:

(NH4)2CO3(s) + H2O(l) ( 2NH4+1(aq) + CO3-2(aq)

This could produce a solution that is .

The ammonium in ammonium carbonate is from the weak base .

So, OH-1 (from the self ionization of water) is attracted to the ion:

NH4+1(aq) + OH-1(aq) ( NH4OH(aq)

Now, since there is less , the self-ionization of water reaction will

shift to the . More H2O will self ionize creating more (which will also get ) and more . The increase in ion makes the solution .

but….

The carbonate in ammonium carbonate is from the weak acid .

So, H+1 (from the self ionization of water) is attracted to the ion:

2 H+1(aq) + CO3-2(aq) ( H2CO3(aq)

Now, since there is less , the self-ionization of water reaction will shift

to the . H2O will self ionize creating more (which will also

get ) and more . The increase in makes the

solution .

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.

Ex4: Hydrolysis of the salt of a strong acid and strong base:

NaCl(s) + H2O(l) ( Na+1(aq) + Cl-1(aq)

The sodium ion will not combine with (from the self ionization

of water). If it did, , a strong base would form. This strong base would immediately dissociate back into and . The following reaction will not occur:

Na+1(aq) + OH-1(aq) ( NaOH(aq)

The chloride ion will not combine with (from the self ionization

of water). If it did, , a strong acid would form. The strong acid would immediately ionize back into and . The following reaction will not occur:

H+1(aq) + Cl-1(aq) ( HCl(aq)

So, this type of hydrolysis could produce a solution that is .

Ex5: What kind of solution would be produced by the hydrolysis of the following salts?

NH4Br K2SO4 CaCrO4 Na2CO3 Fe(C2H3O2)3

Buffer solutions:

Sometimes solutions need to be made “ ” in pH.

An example is . The pH can vary between 7.3 and 7.5 but…

under 6.9 results in = death

above 7.7 results in = death

so… blood has buffers to help keep pH relatively

An example of a buffer solution: HC2H3O2 and NaC2H3O2

Normally:

HC2H3O2 + H2O ↔ H3O+1(aq) + C2H3O2-1(aq)

(slightly acidic)

Normally:

NaC2H3O2 + H2O ( Na+1(aq) + C2H3O2-1(aq)

This is the hydrolysis of the salt of a solution (it is slightly ).

so…. when an like HCl is added to the solution, H+ from the HCl combines

with the from the salt it.

so…. when a like NaOH is added to the solution, OH-1 for the base combine

with from the acetic acid it.

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

Buffers are made from acids and the of that weak acid (buffer in the

range). A buffer can also be made from a weak and its salt (buffer in the

range).

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pH

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