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 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 =
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 Base1 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 Base2
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 à 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
D
Ex2: H2SO4 + 2 NaOH à 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
D
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 / H3PO4Moles 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 à Salt + Water
Ex1: Write the molecular, total ionic, and net ionic equations for the neutralization of
HCl with NaOH.
(molecular) HCl(aq) + NaOH(aq) à NaCl(aq) + H2O(l)
+H2O +H2O
(total ionic) H3O+1(aq) + Cl-1(aq) + Na+1(aq) + OH-1(aq) à Na+1(aq) + Cl-1(aq) + 2H2O(l)
(net ionic) H3O+1(aq) + OH-1(aq) à 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) à 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) à 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) à 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) à 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) à 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) à 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) à 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