Arnoldi
Arnoldi
Honors Chemistry
Acids and Bases Part Two
Note Sheet Answer Key
I. Self-Ionization of Water
A. Background Statements
1. Tap / Natural Water conducts electricity.
Why don’t you swim in a lake during a thunderstorm?
Why does it conduct?
Because it is full of many ______ (i.e. F-1 and Cl-1).
2. Pure water does not contain ions dissolved in it, so it won’t conduct… or does it?
B. Water Self-Ionizes Itself
1. Equation HOH(l) + H2O(l) ( H3O+1 (aq) + OH-1 (aq)
This occurs 0.0000002% in the forward direction and 99.9999998 % in the reverse reaction. Therefore, it makes extremely few ions, and to our eyes using Blinkee, we would not see it conducting.
2. Relationship to Keq
a. Remember, Keq is the __________________.
b. For the self-ionization of water, we call it ___.
]
C. [Ion] Relationship
(i) The value of Kw is ______________.
(ii) In __________ solutions, like pure water, for every one hydronium ion made, there is ________ hydroxide ion also. (at 25’C)
What does this mean mathematically?
(iii) If the solution is acidic…
[H3O+1 ] [ OH-1]
(iv) If the solution is basic…
[H3O+1 ] [ OH-1]
(v) The product is always ____________!
COMPLETE PROMPT ONE NOW
II. Fundamental Acid / Base Math
A. pH, pOH
1. Ranges / Strength Comparison
a. pH
|--------------------|--------------------|
1 7 14
b. pOH
|--------------------|--------------------|
1 7 14
ex: What would 6 be?
c. Truth be told, it is from ____________ since no acids/bases ever get more than 100 M solutions.
2. A Deeper Look at pH
a. Definition
(i) measures PPM of Hydrogen Ion
(ii) logarithmic function of [H+]
Logs are functions of exponents… for example…the log of 1000, which is 103 =
the log of 0.01, which 10-2 =
b. Equation
c. Examples, Given [H+] find pH
(in your calculator: - key, log key, number value, enter key)
If [H+] = 9.5 X 10-10, what is pH? Is this an acid or base?
From an equation, what is the pH?
H2SO4 + H2O ( HSO4 -1 + H3O+1
[0.0002M]
COMPLETE PROMPT TWO NOW
4. A Closer Look at pOH
a. Definition same as pH, except we are talking about the __________
b. Equation
c. Examples, Given [OH-1] find pOH
If [OH-1] = 0.00085, what is pOH? Is this an acid or a base?
If 15.0 g of KOH dissolve in 35.0 L of water, and they fully dissociate, what is pOH?
Mg(OH)2 + H2O ( Mg +2 + 2OH-1 What is pOH?
[0.00675 M]
COMPLETE PROMPT THREE NOW
5. Relationship Between pH and pOH
a. Equation
b. Example
0.059 g of sulfuric acid fully ionizes in 2.00 L of water. What is the pOH of the solution?
COMPLETE PROMPT FOUR NOW
B. [H+] and [OH-]
1. [H+]
a. Definition: concentration of hydrogen / hydronium ions
b. Equation
(on your calculator: 2nd button, log button, - button, pH #, enter)
c. Example
What is the [H+] if pH = 8.9?
2. [OH-]
a. Definition: concentration of hydroxide ions
b. Equation
3. Relationship between [H+] and [OH-]
a. Equation
b. Example
What is the [OH-] of a solution if the [H+] = 0.010?
C. Box
pH + pOH = 14
pH = - log [H+] pOH = - log [OH-]
[H+] = antilog –pH [OH-] = antilog -pOH
[H+][OH-] = 1.00 X 10-14
D. Example
What is the pH, pOH, [H+], and [OH-] if 0.0095 moles of sulfuric
Acid completely ionize in 2.50 L of water?
COMPLETE PROMPT FIVE NOW
III. Titrations
A. Definitions
1. ________: the use of a solution of known concentration to determine
the unknown concentration of another solution
2. ______________: the solution of known concentration in the
titration
3. What is the goal of a titration? to get equal amounts of H+1 and OH-1 ions
NOTE: getting a perfect match of H+1 and OH-1 ions is next to impossible, so most neutral solutions are actually slightly __________ or slightly __________.
4. Graphically speaking, what does this look like?
[pic]
Label: high amount of H+1, H+1 = OH-1, High Amount of OH-1
B. Acid / Base Neutralization
1. General Process:
2. Standard Net Ionic Equation
REVIEW: KOH (aq) + HCl (aq) ( KCl (aq) + HOH (l)
K+1 (aq) + OH-1 (aq) + H+1 (aq) + Cl-1 (aq) ( K+1 (aq) + Cl-1 (aq) + H2O (l)
3. Connection to Acid / Base Indicators
a. Basic Concepts for Indicators
(i) Technically, what is an indicator?
An indicator is a _______ where the color depends on the amount of ____________ present.
An indicator is used to show the ______ of solutions.
(ii) What are indicators made of?
Indicators are usually made up of _________.
Their general formula is:
HIn in water H+1 + In-1
Color One ColorTwo
Acid Color Base Color
(ii) How do indicators work?
In essence, it is the balance between two
sets of opposing forms… it’s Le Chatelier!!!
HIn in water ( H+1 + In-1
HOH(l) + H2O(l) ( H3O+1 (aq) + OH-1 (aq)
In a Basic Solution: BOH in water ( B+ + OH-
When an indicator is added, the ______ from the indicator
reacts with ______ ions from the base to make ________.
(Think _________________ in the second equation above)
Now, all the hydrogen ions on the right side of the indicator equation are _______, which disturbs the __________, and according to Le Chatelier’s principle, the indicator reaction shifts to the ___________to fill in the H+1 “hole”.
This creates more _________, and the color that shows is the __________ color.
HIn in water H+1 + In-1
HOH(l) + H2O(l) ( H3O+1 (aq) + OH-1 (aq)
In an Acidic Solution: HX in water ( H+1 + X-1
When an indicator is added, there is a very high amount of ______.
This disturbs the equilibrium in the _________ equation, and according to Le Chatelier’s principle, the reaction shifts to the left to use up some of the excess H+1 .
This creates more ______, and the color that shows is the ______________
HIn in water H+1 + In-1
HOH(l) + H2O(l) ( H3O+1 (aq) + OH-1 (aq)
(iii) What indicators should you know?
Phenolphthalein
Methyl Orange
Bromothymol Blue
Litmus Paper
COMPLETE PROMPT SIX NOW
b. Definitions for Titrations and Indicators
(i) Transition Interval
pH range over which the indicator ________
this is a ____________ range
(ii) End Point
the point in the titration where the indicator ___________
(iii) Equivalence Point
the point in the titration where [H+] [OH-]
c. Choosing an Appropriate Indicator
Our goal is to get the ____________ as close to the ____________________ as is possible.
Possible Scenarios (Ka, Kb, Solubility Chart)
SA/SB SA/WB WA/WB WA/SB
Which substance has a bigger impact (per drop) on the pH?
Reminder:
Strong Acids to Know:
Strong Bases to Know:
(how is the Ss accounted for? )
Weak Acids to Know:
SA/SB
1 drop of either solution __________________
So, _____________________ will work
i.e.
SA/WB
_____________ has bigger impact
Pick an indicator that changes color _______________.
i.e.
WA/WB
___________ has a big impact
_____ indicator will work well. This is a ____ titration!
WA/SB
___________________ has bigger impact
Pick an indicator that changes color in _____________. i.e.
d. pH Meter for Accuracy
COMPLETE PROMPT SEVEN NOW
C. Mathematics of Titration
1. Equivalents
a. Definition: the number of ____________ions involved in the titration
b. Connection to Titrations:
The titration is complete when the number of H+1 ions ___ the number of OH-1 ions.
Said in another way, the titration is complete when
________ = ________.
Said another way, moles of H+ must equal moles of OH-.
(We hope this occurs near the ___________.)
(what’s that again?)
c. Number of Equivalents Equation
(if the ionization / dissociation is complete, the # of ions = subscript)
Example
How many equivalents react if 9.30 of H2CO3
acid are fully ionized?
What does this means in terms of titrations? It means, we’ll need __________ OH- to titrate (neutralize) this acid.
2. Equivalence
a. Chemical Equivalents: quantities of solutes that have _______________________
in other words: the mass of each substance per ____ or ____ ion
For balance, we need __________ for each _________!
(now… for our purposes… in neutralization, the H+ and OH- completely come together… we don’t need to worry about partial ionization or dissociation)
(i) What does this look like in an equation?
HCl + NaOH ( NaCl + H2O
__ mole of HCl is required to balance __ mole of NaOH.
H2SO4 + 2 NaOH ( Na2SO4 + 2H2O
Only __ mole of H2SO4 is needed to balance __ mole of NaOH.
How many moles of H3PO4 are needed to balance 1 mole of NaOH?
(ii) Equation:
(another way of saying the last equation)
Example: How many grams of H2CO3 would equal 0.290 equivalents?
b. Gram Equivalent Masses
(i). Definition: the number of grams of an acid or base that will provide ______ of protons (___ ions) or _______ ions
(ii). Equation
Molecular Mass is from
equivalents = number of ions involved
___ for acid
___ for base
Examples
What is the equivalent mass of H2SO4? Assume full ionization.
What is the equivalent mass of H2SO4 from this ionization equation?
(which might occur if there was only
partial ionization)
H2SO4 + H2O ( HSO4-1(aq) + H3O+1(aq)
COMPLETE PROMPT EIGHT NOW
3. Normality
a. Definition: a concentration unit
Normality is important for acid/base reactions because it always accounts for the _________ (H+1 / OH-1) involved
(thus, it is better than molarity)
b. Equations
aka
(assuming subscript = ions involved)
aka
c. Example
0.00475 g phosphoric acid completely ionizes in 15 L
water. What is the molarity of the solution?
What is the normality of the solution?
4. Actual Titrations Math
a. What happens if we mix an acid and a base, and do NOT use equivalent amounts of each?
Example: Find the pH of a solution made by mixing 50.0 mL of .100 M HCl with 49.0 mL of .100 M NaOH.
FIRST: Find the moles of HCl and moles of NaOH:
SECOND: Find the equivalents of H+ in HCl and the equivalents of OH- in NaOH.
THIRD: Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+ and equivalents of OH- from each other (absolute value). 0.0010 eq H+
FOURTH: Calculate the normality of the resulting solution:
NOTE: TOTAL LITERS!!!
FINALLY: Calculate the pH.
(where we used M of H+ before, now use N)
Example: Find the pH of a solution made by mixing 50.0 mL of .100 M H2SO4 with 50.0 mL of 1.00 M NaOH.
FIRST: Find the moles of H2SO4 and moles of NaOH:
SECOND: Find the equivalents of H+ and the equivalents of OH-
THIRD: Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+ and equivalents of OH- from each other (absolute value).
FOURTH: Calculate the normality of the resulting solution:
FINALLY: Calculate the pOH and convert to pH.
b. How do we determine either:
(i) the exact amount of acid / base we need to titrate?
(ii) the unknown concentration based upon the volumes we used to titrate?
EXAMPLE: 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?
First, calculate the moles of KOH:
.215 M = x / .0155 L ( x = .00333 moles KOH
Second use “stoich” to convert moles of KOH to moles of HC2H3O2:
HC2H3O2 + KOH ( KC2H3O2 + HOH
.00333 moles KOH x 1 moles HC2H3O2 x = .00333 moles HC2H3O2
1 moles KOH needed to cancel out
(completely use KOH)
Last, calculate the molarity of the acid:
M = .00333 moles / .0212 L ( x = .157 M
THAT’S THE HARD WAY!!!!
LET’S DO IT THE EASY WAY…
V =
M =
# =
a =
b =
V =
N =
a =
b =
Examples
(i) What volume of a 1.0 M solution KOH is needed
to titrate 13 mL of a 0.15 M solution of phosphoric
acid?
(ii) 145 mL of calcium hydroxide were required to
titrate 130 mL of a 2.50 N solution of nitric acid. What is the base normality?
More Examples… uh-oh!!! THINK!!!
If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid, calculate the % acetic acid in solution.
First, use the titration formula to find the normality of the acid.
NaVa =NbVb
(x)(20.30 mL) = (.750 N)(18.75 mL) ( x = .693 N
Then, multiply the normality by the equivalent weight of the acid.
.693 N ( .693 eq/L
(.693 eq)(60.0g) = 41.6 g/L
L 1 eq
Third, convert the liters to grams: (It’s water!)
41.6 g ( 41.6 g ( 41.6 g
L 1000mL 1000g
Finally, calculate the % by mass:
g acid X 100 = (41.6 g) X 100 = 4.16 %
g solution 1000g
That was the difficult way. The easy way …
If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid, calculate the % acetic acid in solution.
FIRST: Use the titration formula to find the normality of the acid.
SECOND: Use this equation.
Sorry, there is no easy way around this one…
How many mL of a 1.20 % HCl solution are needed to titrate 25.50 mL of .100 M magnesium hydroxide?
FIRST: What does 1.20 % mean?
SECOND: Multiply everything by 10 to get the amount of solute in 1000 grams
(which is 1000 mL, which is 1 L).
THIRD: Find molarity by converting grams of solute to moles.
FINALLY: Use the titration formula to find the mL of the acid.
COMPLETE PROMPT NINE NOW
IV. Salts in Solution
A. Basic Ideas
________________: the reaction of a substance with water.
a. Acids make ______________ solutions.
b. Bases make ______________ solutions.
c. Salts sometimes make ____________ solutions.
However, they sometimes lead to _______________ solutions.
Hmmm…
B. EXAMPLES:
Hydrolysis of the salt of a strong base and weak acid:
First, remember that water self-ionizes. Therefore, in aqueous solutions, this is happening:
H2O(l) + H2O(l) ( H+1(aq) + OH-1(aq)
Now, if I put sodium acetate, a soluble salt in water, it will dissociate like this:
The sodium in sodium acetate could combine with the hydroxide ions to form the base ______________. However, since NaOH is a ______ base, there is no attraction of sodium ions for any of the hydroxide ions present from the self ionization of the water.
but….
The acetate in sodium acetate could combine with the hydronium / hydrogen ions to form acetic acid. Now, acetic acid is a weak acid, so there is an equilibrium that can be disturbed and this does occur. The H+1 (from the self ionization of water) ___________ to the acetate ion, and thus forms the following:
(not all, just some)
Of course, this disturbs the self-ionization of water equilibrium. Because there is less _____, the self-ionization of water reaction will _____________. This fills in the H+ “hole”, but adds an excessive amount of ________ which results in a ____________________.
Hydrolysis of the salt of a strong acid and weak base:
First, don’t forget water is self-ionizing.
H2O(l) + H2O(l) ( H+1(aq) + OH-1(aq)
Now, if I put ammonium chloride, a soluble salt in water, it will dissociate as follows:
The chlorine in ammonium chloride could combine with the hydronium / hydrogen ions in the self-ionization equilibrium to form HCl. However, since HCl is a __________ acid, this does not occur.
but….
The ammonium in ammonium chloride can combine with the hydroxide ion to form the _______ base, ammonium hydroxide. When the OH-1 (from the self-ionization of water) is attracted to the ammonium ion this occurs:
Of course, this used up some of the OH-1 ions, so, sthe self-ionization of water reaction will shift to the right. More H2O will self-ionize to fill in the OH-1 “hole”. When this occurs, we get excessive amounts of _____. The increase in hydrogen ion makes the solution ____________.
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 ____________________.
WHY?
The ammonium in ammonium carbonate can combine with the OH-1 ions to form the _____ base NH4OH as follows:
Of course, since there is less OH-1, the self-ionization of water reaction will ______________ to fill in the OH-1 hole, but also making more H+1. The increase in hydrogen ion makes the solution _________________.
But at the same time….
The carbonate in ammonium carbonate can combine with the hydrogen / hydronium ions to form the weak acid HCO3-1 like this:
Which can combine again to form the weak acid H2CO3-1 like this:
Of course, since there is less H+1, the self-ionization of water reaction will _______________ to fill in the H+ hole which also makes more OH-1. The increase in hydroxide makes the solution ____________.
SO WHICH IS IT? ACIDIC or BASIC?
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:
NaCl(s) + H2O(l) ( Na+1(aq) + Cl-1(aq)
The sodium ion will not combine with hydroxide (from the self- ionization of water) because it would form the strong base, NaOH which would immediately ________ back into sodium and hydroxide ions. So, the following reaction will _____ occur:
Na+1(aq) + OH-1(aq) ( NaOH(aq)
The chloride ion will not combine with H+ (from the self ionization of water) because it would form the strong bae, HCl which would immediately _________ back into H+ and Cl-1. So, the following reaction will ___________ occur:
H+1(aq) + Cl-1(aq) ( HCl(aq)
Therefore, there are no equilibrium shifts to the self-ionization of water, no extra hydrogen / hydronium ions or hydroxide ions are made, and the solution stays ______________.
PUTTING IT ALL TOGETHER
What kind of solution would be produced by the hydrolysis of the following salts?
Note: I’m not asking why, I just asked what happened.
Basically, I’m asking if the positive ion is from a ________________, and if the negative ion is from a _____________________.
What did we learn above?
___________________ result in making bases, which cause an OH-1 hole, so it shifts making more H+, and we end up with an ___________________.
____________________ result in making acids, which cause a H+1 hole, so it shifts making more OH-1, and we end up with a
____________________.
_________________________________ together could go either way, and we ___________________.
______________________________ make nothing, cause no holes, cause no shifts, and result in _____________ solutions.
So…
NH4Br K2SO4 CaCrO4 Na2CO3 Fe(C2H3O2)3
COMPLETE PROMPT TEN NOW
C. Buffer Solutions
1. Sometimes solutions need to be made to be “____________
____________”.
2. An example is blood. The pH can vary between 7.3 and 7.5 but…under 6.9 results in acidosis = death, and above 7.7 results in alkalosis = death.
so… blood has buffers to help keep pH relatively constant.
3. An example of a buffer solution: HC2H3O2 and NaC2H3O2
Normally:
HC2H3O2 + H2 O ( H3O+1(aq) + C2H3O2-1(aq)
This is acetic acid, a ____________acid.
NaC2H3O2 + H2O ( Na+1(aq) + C2H3O2-1(aq)
This is the hydrolysis of the salt of a Strong Base /Weak Acid solution and it is _______________.
Now, if we introduce an acid or base to the solution, these “buffers” will act in a way to ________________ _________so that the pH does not change significantly.
For example, if an acid like HCl is added to the solution…
The extra H+ from the HCl combines with the acetate from the salt neutralizing the HCl effect.
HC2H3O2 + H2 O ( H3O+1(aq) + C2H3O2-1(aq)
If a base like NaOH is added to the solution…
HC2H3O2 + H2 O ( H3O+1(aq) + C2H3O2-1(aq)
NaC2H3O2 + H2O ( Na+1(aq) + C2H3O2-1(aq)
.
OH-1 for the base combines with H+ from the acetic acid
H2O(l) + H2O(l) ( H+1(aq) + OH-1(aq)
neutralizing the KOH effect.
so…. this buffer solution is resistant to change in pH.
4. Buffers are made from weak acids and the salt of that weak acid (buffer in the acid range).
OR
A buffer can also be made from a weak base and its salt (buffer in the base range).
COMPLETE PROMPT ELEVEN NOW
-----------------------
pOH
pH
[OH-]
[H+]
................
................
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