(Cj) Chapter 10: Acids And Bases

Front 1

Arrhenius Definition

Back 1

According to this model, an acid is species that will give off protons and a base is a species that will give off OH ions

Front 2

Bronsted Lowry Definition

Back 2

According to this model, an acid is a species that donates protons and a base is a species that will accept protons

Front 3

Conjugate Acid and Base Pairs

Back 3

According to the Bronstend Lowry Model, acids and bases will always occur in pairs. Once the acid gives up its proton will will turn into a species that can not accept protons making it a base. Once an acid give up it proton it becomes its conjugate base and vice versa

Front 4

Lewis Definition

Back 4

A lewis acid is a species that accepts electrons while a lewis base is a species that will donate electrons

Front 5

Naming Anions

Back 5

If an anion ended with the letters -ide then when it becomes an acid it takes on the prefix hydro- and the suffix -ic acid. For example, A Fluoride anion will take on a hydrogen and become hydrofloric acid

Front 6

Naming Oxyanions

Back 6

If the oxyanion ended with the suffix -ite (as in the case of hypochlorite) then when it takes on a hydrogen the suffix changes to -ous (as in the case of hypochlorous acid). If the oxyanion ended with the suffix - ate (as in the case of nitrate) when it takes one the hydrogen the suffix becomes -ic (as in the case of nitric acid). Note: ate ic ite ous

Front 7

Amphoteric

Back 7

As in the case of water, when this type of species is exposed to an acid it acts like a base and when it is exposed to a base it acts like an acid

Front 8

Auto Ionization

Back 8

Water has the ability to react with itself to form a hydronium ion and hydroxide ion. It does this by donating a proton to another water molecule. The molecule that donated the proton becomes a hydroxide ions and the molecule that accepted the protons becomes a hydronium ion.

Front 9

Kw = Water Dissociation Constant

Back 9

Kw = {H30}{OH} = 10 ^ -14 at 298 kelvin

Front 10

When water ionizes, what will the concentration be of the individual ions?

Back 10

Based on the water dissociation constant, the concentration will be 10 ^ -7 for each ion. But this is only in the case where the water is alone.

Front 11

When will the value for Kw change?

Back 11

At 298 kelvin the value for K will never change. It is a constant at that given temperature so it will not change unless the temperature has changed.

Front 12

Equation for pH

Back 12

pH = -log {H+}

Front 13

Equation of pOH

Back 13

pOH = -log{OH-}

Front 14

If you have pH, how would you find pOH?

Back 14

Simply subtract from 14.

Front 15

Aside from negative using the -log(Ka) method, how do you convert a Ka to a pKa?

Back 15

Take the scientific notation of the Ka expression. Turn the exponent into an actual and subtract from that the -log(x) of what ever number was in the front. How do you deal with that negative log? The value is going to be something between 0 and 1 (remember that log 10 = 1). So if the number in the log is small, then after the subtraction the final number will be closer to the number that was the exponent than one minus that value.

Front 16

Weak Acid and Bases and strong acids and bases

Back 16

When we talk about weak vs. strong acids or bases we are talking about the tendency of that given acid or base to dissociate in a given solvent. A strong acid will dissociate to a much further extent than a weak acid and will be much more concentrated.

Front 17

Will simply taking the molar concentration of an acid or base be enough to find the pH?

Back 17

No. If the acid is in water, you also need to factor in the presence of hydrogen ions that dissociated from the water. In the case of 1 x 10 ^ -8 M solution of HCl you need to take that value and place it in the water ionization constant to find the effect of the water.

Front 18

Size of Ka

Back 18

The smaller the Ka the weaker the acid. Ka tells you the extent to which the acid will dissociate so if its small it means that it will only dissociate to a limited extent.

Front 19

Expression for Ka

Back 19

Ka = {H30}{A-}/{HA}

Front 20

Expression for Kb

Back 20

Identical for the expression for Ka

Front 21

Ka and Kb of conjugate acid base pair

Back 21

If you have one concentration of a conjugate acid base pair then you can find the other since the product of the Ka of an acid and the Kb of the conjugate base equals the water ionization constant. This is given by the equation Ka (conjugate acid) x Kb (conjugate base) = Kw = 10 ^ -14

Front 22

If you are told that you have a 2 Mol solution of acetic acid with a Ka of 1.8 x 10 ^ -5, how would you find the hydrogen ion concentration?

Back 22

1) Write you the equation for the dissociation
2) The equation for Ka is {H30}{A-}/{HA).
3) You know the value for Ka and the value for HA. All you need to do is solve for X
4) There is a simplifying assumption in which you remove the -X from the bottom that you will usually do to make the calculation simpler.

Front 23

Neutralization Reaction

Back 23

Reaction between an acid and a base

Front 24

Hydrolysis

Back 24

When a salt reacts with water to give back the ions from which it is composed

Front 25

Polyvalence

Back 25

When an acid gives over a proton, it is said to give up one equivalent weight of protons or hydroxide ions if it is a base. HCl for example will give up one mole of hydrogen protons for every mole of HCl. Polyvalence occurs when an acid can give up more than one mole because it have more than one hydrogen. One mole of H2SO4 can give you two moles of hydrogen since it have two hydrogens to give up.

Front 26

Titration

Back 26

A procedure used to find the molarity of a known reactant in a solution. I titrrant of known volume and concentration is added to a solution where the volume may be known by the concentration may not be. An indicator is used to tell at which point the number of mole of titrant is equivalent to the number of moles of titrand

Front 27

Equivalence point

Back 27

This is the point where the number of mole of titrant is equivalent to number of mole of titrand

Front 28

What is the difference between the end point and the equivalence point?

Back 28

The end point is the point at which the indicator changes color and the equivalence point is the point at which the number of moles of titrant equals the number of moles of titrand. These two point are generally close enough that the difference can be ignored.

Front 29

Where will the equivalence point be for a reaction between a strong base and a strong acid?

Back 29

The equivalence point will be at a pH of 7. Use the tug of war analogy to estimate where the equivalence point will be for a reaction between acids and bases of various strengths.

Front 30

How can you tell what type of species are being used based on the acid base curve?

Back 30

Simply look at the where the curve starts out. Does it start in a more acidic range or a more basic range? You can also tell based on the equivalence point what types of species were used in the reaction.

Front 31

Henderson Hasselbach Equation

Back 31

This equation is used to determine the amount of acid and base present in the buffer region of a solution.

Front 32

how can you estimate a log
n x 10-m =

Back 32

log(n x 10-m)= -m + logn
the negative log is
m-logn

Front 33

m-logn
since n is a number betw 1 and 10 its log will be a fractoin betw ?

Back 33

0 and 1

Front 34

If Ka = 1 .8 x 10-5, then pKa =
estimate please

Back 34

5 - log 1.8. Since 1 .8 is small,
its log will be small, and the answer will be closer to 5 than
to 4. (The actual answer is 4.74.)

Front 35

13. Which of the following combinations would produce a buffer solution
of pH = 4?

(Ka HN02=4.5 X 10-4)
A. 0.30 M HNO2, 0.22 M NaNO2
8. 0.22 M HNO2, 0.30 M NaNO2
C. 0.11 M HNO2, 0.50 M NaNO2
D. 0.50 M HNO2, 0.11 M NaNO2

Back 35

pH = PKa + log [A-/[HA]
4 = 3.35 + log [A-]/[HA]
0.65 = log [A-]/[HA]
[A-]/[HA] = 4.5
Only Choice C fulfills this criterion as 0.50/0.11 = 4.5.

Front 36

Blood pH is maintained in a relatively small range (slightly
above 7) by a

Back 36

bicarbonate buffer
system. This homeostasis can be upset, leading to a condition known as acidosis.

Front 37

henderson-hasselbach
pH = PKa + log [conjugate base]/[weak acid]
when will the concentr of conj base equals the conc of the weak acid ?

Back 37

in a titration, half-way to the equivalent point

Front 38

henderson-hasselbach
pH = PKa + log [conjugate base]/[weak acid]
when the concentr of conj base equals the conc of the weak acid then?

Back 38

ph=pKa, bec log1=0

Front 39

the henderson-Hasselbalch equation is used for?

Back 39

The Henderson-Hasselbaich equation is used to estimate the pH ot
a solution in the buffer region where the concentrations of the species
and its conjugate are present in approximately equal concentrations.

Front 40

Consider a buffer solution of acetic acid and sodium acetate:

what happens when a small amount of HCl is added?

Back 40

H+ ions from the HCl react with the acetate ions to form acetic acid. Thus
[H+] is kept relatively constant and the pH of the solution is relatively
unchanged.

Front 41

Consider a buffer solution of acetic acid and sodium acetate:

CH3COOH <---> H+ + CH3COO-
what happens when NaOH is added to the buffer?

Back 41

When a small amount of NaOH is added to the buffer, the OH- ions
from the NaOH react with the H+ ions present in the solution; subse-
quently, more acetic acid dissociates (equilibrium shifts to the right),
restoring the [H+]. Thus, an increase in [OH-] does not appreciably
change pH.

Front 42

what formula do yo use to calculate the volume added to reach the endpoint?

Back 42

NaVa=NbVb

Front 43

what is equival weight of H2SO4?

Back 43

98 g/mol is molec weight
since each mole liberates 2 acid equiv, the gram equiv of H2SO4 is 98/2 or 49 g.

Front 44

how do you calcul equival weight?

Back 44

divide the gram molecular weight by how many moles of H+ it liberates

Front 45

the acid HClO has a Ka = 3.2 X 10-8,
and the base NH3 has a Kb = 1.8 x 1O-5.
what will the resulting solut be? acidic, basic, or neutral?

Back 45

an aqueous solution of HCIO
and NH3 is basic since Ka for HCIO is less than Kb for NH3

Front 46

what are the react for HCL reacting with NH3?

Back 46

HCl (aq) + NH3 (aq) --> NH4+ (aq) + Cl-(aq) Reaction I

NH4+ (aq) + H20 (aq) --> NH3 (aq) + H3O+(aq) Reaction II

Front 47

Ka= [X][X]/[2.0]=1.8 X 10-5

what did you approximate?

Back 47

that 2.0-x is approx = to 2.0

Front 48

Ka= [X][X]/[2.0-X]=1.8 X 10-5

if when you solve for X it is close the the original concentr of acetic acid (2.0 M) what do you need to do?

Back 48

you cant approx that 2.0-x is =2.0
instead you have to use the quadratic equation

Front 49

strong acids commonly encountered include (6)

Back 49

HCl
HBr
H2SO4
HNO3
HI
HClO4

Front 50

combinations of acids/bases
1. HCl + NaOH --> NaCl + H2O

2. HCl + NH3 --> NH4Cl

3. HClO + NaOH --> NaClO + H2O

4. HClO + NH3 --> NH4ClO

Back 50

SA + SB
SA + WB
WA + SB
WA + WB