FINAL Chapter 15/16/17

Question 1

CHAPTER 15
a
Let’s say initally the equilibrium was

(a) + (b) = (c)

K = (c)/(a)(b) and then the volume is doubled. Which direction would the reaction shift?

Answer 1

Reaction would shift left. Volume is doubled meaning all pressures are halved soooo

Q= 1/2(c)/1/2(a)1/2(b)

with the fractions that means overall you end up with 4/2 = 2

so Q = 2K meaning that Q is greater than K and the equilibrium must shift left

Question 2

CHAPTER 17

To an aqueous solution originally containing 1.0 x 10-4 mol/L Ag+ was
added enough of the weak acid HCN to reach an equilibrium concentration of [HCN] = 3.0 x 10-4
mol/L. What must be the pH of the solution in order to observe a precipitate of AgCN?
(Hint: Assume any change in pH will not change the value of [HCN]).
𝑲𝒔𝒑 𝒐𝒇 𝑨𝒈𝑪𝑵 = 𝟏. 𝟐 × 𝟏𝟎−𝟏𝟔
𝑲𝒂 𝒐𝒇 𝑯𝑪𝑵 = 𝟔. 𝟐 × 𝟏𝟎−𝟏

Answer 2

1. Write out both the equilibrium constant expression and the dissolution constant expression.
2. At equilibrium Q = Ksp so you find the CN- value which makes this so

Q = [Ag+][CN-]
This CN- value is at equilibrium

3. Then you sub in this value into the Ka expression to find the H3O+ concentration
4. Following this you take the log to find the pH, and in order for precipitation to occur, the pH must be higher than this value (becuase CN is a base so higher pH means higher concentration so teh product of Ag and CN will be more than the Ksp

Question 3

CHAPTER 15
What is the only factor that k changes?

Does LeChatlier’s changes affect k?

Answer 3

Temperature and it changes proportionally

No, only if temperature

Question 4

Define equilibrium

Answer 4

Rate of forward and reverse reactions are equal, concentraction of reactant and product are constant

Question 5

Define activity

Answer 5

Effective concentrations or pressures for real gases factoring in particle interactions

Question 6

How do teh following changes to the reaction affect k?

1) Reversing
2) Doubling coefficents
3) Adding two reactions
4) Subtracting two reactions

Answer 6

1) Reciprocal
2) Squaring
3) Multiplying K’s
4) Dividing two K’s

Question 7

With le Chatlier’s, what’s two trick to watch out for?

Answer 7

Changing the amount of concentraction of solids of liquids does not change the equilibrium, unless it is the solvent

Catalyst affects forward and reverse reaction rates equally so no effect on equilibrium

Question 8

Acid-Base Definitions

Arrenenius
Bronsted-Lowry
Lewis

Answer 8

Arrenenius
- Acid (produces H)
- Base (produces OH)
Bronsted-Lowry
- Acid (proton donor)
- Base (proton acceptor)
Lewis
- Acid (electron pair acceptor)
- Base (electron pair donor)

Question 9

Strong and weak acids, what expressions hold true?

Answer 9

Strong: [H+] = [HA]0
Weak: [H+] &laquo_space;[HA]0

Question 10

pKa calcualtion and what low and high pKa mean

two other important equations

Answer 10

pka = -log(Ka)

low pka means strong
high pka means weak

[H+] = 10^-pH
Ka x Kb = Kw

Question 11

CHPATER 18

Buffer region

And buffer

Answer 11

Half equivalence point where concetration of acid equals that of conjugate base of pH = pKa

Resist changes in pH when small amounts of acids or base are added

Approx equal but high concentrations of weak acid and conjugate base or weak base and conjugate acid

Question 12

Two ways to prepare a buffer:

Answer 12

1) Equal moles of acid and its conjugate base

0.1mol A + 0.1mol B in 1L equals 0.2mol/L

2) Partial neutralization: weak acid and strong base to generate the conjugate base

0.2 mol A and 0.1 mol B (more acid than base) to make 0.2 mol/L

Question 13

CHAPTER 18

When can you neglect the x value to simplify quadratic?

x + 0.05

Answer 13

For the dissociation of a weak acid, where the change in concentration is not significant

So long as the x/10^-7 > 100

You would neglect the 0.05 for very strong acids or bases where the inital concentration is so small to the change

Question 14

What is teh common ion effect in relation to solubility

Answer 14

The addition of soluble salt that shares a common ion with insoluble salt reduces solubility of insoluble salt

Question 15

When does precipitation occur?

What does Q = Ksp mean/

Answer 15

When Q is greater than Ksp becuase the solution is beyond saturation

System as at equilibrium

Question 16

when does pH affect the solubility of salts

Answer 16

when anion is the conjugate base of a weak acid, because in acidic conditions the conjugate base reacts with H+ to form weak acid, shiftning equilibrium away and increases solubility

Question 17

When will lower Ksp percipitae?

Answer 17

At lower concentration of precipitating ion

Question 18

Find molar solubility of AgCl in 0.1M NH3(aq)

Answer 18

1. Create expression for Ka and for Ksp
2. Add both equations and multiply K’s to get overall K
3. Solve the ice table for S

Question 19

If you have 0.1 + S, when do yuo neglect the 0.1 and the S

Answer 19

You neglect the solubility if it is very small compared to the inita lconcentration of the ion (such as with low ksp)

You neglect inital concentration if it is very small compared to the solubility

Question 20

CHAPTER 16

Calculate the pH of a buffer produced by mixing 50 mL of a 0.1M solution of sodium acetate
with 38 mL of a 0.05M HCl solution. Assume that volumes are additive.
(Kb of the acetate ion is 5.9 x 10-10).

Answer 20

1) Find the moles of the sodium acetate and HCl to deduce which is the limiting reagent

2) Now calculate how many moles of the acid and conjugate base are left. In this case we had

CH₃COO- +HCl→CH₃COOH+Cl-
conjugate base acid
0.005 0.0019

So the acid will have the same moles as the limiting reagent and the moles remaining after the buffer of the conjugate base will be 0.005-0.0019 = 0.0031

If one of these had a coefficient of 2, you have to factor that in!

3) Now calculate the concentrations of both of these, remembering that the volume is additive so it’s 50 + 38 and it’s in L not mL

4) Now to find the pka, first find the kA using Ka + Kb = Kw and then convert to pka

5) Insert all of this into the HH equation ensuring that conjugate base
0.0031/0.0019
Except in concentrations altho in this case it cancels out

Question 21

CHAPTER 17

Assume that you have 1.00 L of a solution that contains 2.50 x 10^-5 M Fe2+ and you saturate
the solution with H2S (equilibrium [H2S] = 0.10 M). What must be the pH of the solution in order to
precipitate FeS?

Ksp of FeS = 3.7 x 10^-19
For the reaction H2S + 2H2O ⇌ 2H3O+ + S2-
, K = 1.1 x 10-19

Answer 21

Solution perciptates when Q is greater than Ksp. Find when Q = Ksp

Q = Ksp = [S2-][Fe2+]
Solve for [S2-] which would also be equilibrium concentration

K = [S2-][H30+]^2/[H2S]
Now isolate and solve for [H30+]^2, square root answer and then input into equation to solve for pH

Question 22

CHAPTER 18

A 𝟎. 𝟐𝟎 𝑴 aqueous solution of some weak acid (𝑯𝑨) has a 𝒑𝑯 𝒐𝒇 𝟓. 𝟏𝟒 at 𝟐𝟓 °𝑪.
What is the 𝜟𝑮° for the ionization of this acid?

Answer 22

1) Use the pH formula to find the [H+], you should get 7.24x10^-6
2) You need to find K and you can do this by setting up an ICE table, and solving for the Ka that way!

you should get

2.62 x 10^-10 when you solve

3) Then you want to input into G = -RTlnk solve for G and you should get 57596 J

Question 23

Under what conditions would it be true to say that as temperature increases, the equilibrium constant for a reaction
increases? Explain

Answer 23

Using the equation

ln(K1/k2) = ∆H/R(1/T2-1/T1)

K2 has to be bigger than K1 so K1/K2 <1 and lnK1/K2 < 0 (less than 0)

If T2 is bigger than T1, than the temperature expression will be negative. In order to ensure the left side is negative as established ∆H has to be positive, meaning an endothermic reaction

Question 24

CHAPTER 16

A buffer is prepared by adding 100 mL of 0.5 mol L-1 NaOH to 200 mL of 0.75 mol L-1 acetic acid (AcOH). What amount
(mol) of HCl must be added to this buffer solution to change the pH by 0.14 units? If necessary, assume the total
volume remains unchanged at 300 mL.

Answer 24

Now let’s make this bad boy a tad easier

First, you want to calculate the moles of AcOH and NaOH (you should get 0.15mol and 0.05mol)

NaOH + AcOH forms A- so the new moles of A- are going to be the moles of NaOH (being the strong base) so 0.05 and the new moles of AcOH will be 0.15 - 0.05 = 0.10

Now in the reaction there is A- and you add acid (HCl) to form HA again (the AcOH) so you will get

A− →HA.

Let x be the new moles of HCl added so new A- will be 0.05 - x and new HA will be 0.10 + x

∆pH = final - inital
-0.14 = the ones with the x’s - the one without the x’s

You can also try with concentrations

Question 25

Explain how the relationship ΔG°= –RTlnK can be used to experimentally determine the values of ΔH and ΔS for a given equilibrium reaction. Include what information would need to be collected and what calculations would need to be performed.

Answer 25

∆G = -RTlnK and ∆G = ∆H - T∆S Set them equal to each other and isolate for lnK lnK = -∆H/R(1/T) + ∆S/R Experimentally you should measure K at different temperatures and plot lnK divided by 1/T. THe slope is -∆H/R and the intercept is ∆S/R