# CHEM126 – CHAPTER 15 – REGULAR HOMEWORK / INTRO ACIDS

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### Question 1

What is the pH of a solution where the [H+] =6.91 x 10-3 M?

REPORT pH TO 2 DECIMAL PLACES. pH=-log(6.91e-3)

pH=2.16

### Question 2

Is the above pH acidic or basic?

Select one: ### Question 3

What is the [H+] concentration if the pH is 7.19? [H+]=10^-7.19

[H+]= 1e^-19

### Question 4

Is the above pH acidic or basic?

Select one: ### Question 5

What is the [H+] concentration if the pOH is 8.19? pOH + pH =14

14-8.19= pH

pH= 5.81

10^-5.81= [H+]

1.549e-6=[H+]

### Question 6

Is the above pH acidic or basic?

Select one: ### Question 7

What is the [H+] concentration if the [OH-] concentration is 9.90 x 10-11 M? ### Explanation

[OH-][H+]=1e-14

[H+]= (1e-14)/(9.9e-11)

[H+]=1.01e-4

### Question 8

Is the above [H+] concentration acidic or basic?

Select one: ### Question 9

Calculate the pH of 0.095 M HClO4. pH=-log[0.095]

pH=1.022

### Question 10

Calculate the pH of 0.618 M HNO3. pH=-log(0.618)

pH=0.209

### Question 11

Calculate the pH of 0.0084 M NaOH. pH=-log(0.0084)

pH=2.076

14-2.076=pH

pH=11.92

### Question 12

Calculate the pH of 0.0019 M Ba(OH)2. Careful. ### Explanation

pOH=-log(0.0019*2)

pH=2.42

14-2.42=pH

pH=11.58

### Question 13

Calculate the pOH of 0.020 M LiOH. pOH=-log(0.02)

pOH=1.69897

### Question 14

Calculate the pOH of 0.17 M HI. pH=-log(0.17)

pH=0.7696

14-pH=pOH

pOH=13.23

### Question 15

Calculate the pOH of 0.00039 M Sr(OH)2. ### Explanation

pOH=-log(0.00039*2)

pOH=3.108

### Question 16

When 1.00 g of thiamine hydrochloride (also called vitamin B1 hydrochloride) is dissolved in water, then diluted to 10.00 mL, the pH of the resulting solution is 4.50. The formula weight of thiamine hydrochloride is 337.3 g/mol. Calculate Ka. ### Explanation

1 g * 1 mol/ 337.3 g = 0.00296472

0.00296/0.01 L = 0.296 M

pH= 4.50 = -log[H+]

[H+]=0.00003

Ka = [H+][A-]/[HA]

Ka= 0.00003^2/0.296

### Question 17

When 0.0021 moles of a weak acid are dissolved in enough water to make 0.500 L, the pH of the resulting solution is 6.04. Calculate Ka. ### Explanation

0.0021/0.5L = 0.0042M [HA]

pH=6.04= -log[H+]

[H+]=9.12e-7

Ka=[H+][A-]/[HA]

Ka= (9.12e-7)^2/(0.0042)

### Question 18

Rank the following acids from least acidic (highest pH) to most acidic (lowest pH). Scheme: Least acidic 5 to most acidic 1. acetic acidChoose…3421-most5-least chlorous acidChoose…3421-most5-least hydrocyanic acidChoose…3421-most5-least hydrosulfuric acidChoose…3421-most5-least phenolChoose…3421-most5-least ### Question 19

Rank the following acids in order of increasing acid strength. Key: bond strengthHFChoose…4-least1-greatest32 HClChoose…4-least1-greatest32 HBrChoose…4-least1-greatest32 HIChoose…4-least1-greatest32 Hint, for similar compounds, does bond strength decrease or increase with increasing size (going down a column)?

### Question 20

Rank the following acids in order of increasing acid strength. Key: Weakening of hydrogen bond and stability of resulting anion. (Electronegativities I=”2.5,” Br=”2.8,” Cl=”3,” and F=”4.)

HOF (if it existed) HOCl HOBr HOI ### Question 21

Rank the following acids in order of increasing acid strength. Key: Stability of anion.HOClChoose…31-greatest4-least2 HOClOChoose…31-greatest4-least2 HOClO2Choose…31-greatest4-least2 HOClO3Choose…31-greatest4-least2 Spreading out the electron density in the anion increases its stability.

### Question 22

In the following acid-base reaction, identify the compounds as acid, base, conjugate acid, and conjugate base.

NH3 + H2O ↔ NH4+ +OH-NH3Choose…conjugate acidconjugate basebaseacid H2OChoose…conjugate acidconjugate basebaseacid NH4+Choose…conjugate acidconjugate basebaseacid OH-Choose…conjugate acidconjugate basebaseacid ### Question 23

For the above reaction indicate the type

Select one: ### Question 24

In the following acid-base reaction, identify the compounds as acid, base, conjugate acid, and conjugate base.

HSO4- + H2O ↔ SO42- +H3O+HSO4-Choose…conjugate acidbaseacidconjugate base H2OChoose…conjugate acidbaseacidconjugate base SO42-Choose…conjugate acidbaseacidconjugate base H3O+Choose…conjugate acidbaseacidconjugate base ### Question 25

For the above reaction indicate the type

Select one: ### Question 26

In the following Lewis acid-base reaction for liquid ammonia, identify the compounds as acid, base, conjugate acid, and conjugate base.

NH3 + NH3 ↔ NH4+ +NH2-

NH3 NH4+ NH2- Note, this is like the autoionization of water.

### Question 27

For the above reaction indicate the type

Select one: ### Question 28

Which of the following is/are correct for the NH2- ion?

Select one or more:  An extremely weak acid is normally not considered an acid. Similar an extremely weak base is normally not considered a base.

### Question 29

In the following Lewis acid-base reaction, identify the compounds as acid, base, conjugate acid, and conjugate base.

Zn(OH)3-(aq) + OH-(aq) ↔ Zn(OH)42-(aq)

Zn(OH)3-(aq) OH- Zn(OH)42- ### Question 30

For the above reaction indicate the type

Select one: ### Question 31

Rank the following species in order of increasing basicity. NH3, NH4+,+NH2-, Cl-,H2O

NH3 NH4+ NH2- Cl- H2O ### Question 32

Which of the following is/are correct for the F- ion?

Select one or more:  ### Question 33

Which of the following is/are correct for the Br- ion?

Select one or more:  ### Question 34

Which of the following is/are correct for the Ca2+ ion?

Select one or more:  ### Question 35

Which of the following is/are correct for the Fe3+ ion?

Select one or more: ### Question 36

Which of the following is/are correct for CH3NH2?

Select one or more:  ### Question 37

Which of the following is/are correct for the H2PO4- ion?

Select one or more:  ### Question 38

Phenol (C6H5OH, also called carbolic acid) has a pKa of 9.89. It is used to preserve body tissues and is quite toxic. Calculate the percent ionized in 0.0868 M phenol–a dilute solution. ### Explanation

pKa= -log(Ka)

Ka= 10^-9.89=1.288e^-10

1.288e-10=[H+][A-]/[HA]=x^2/0.0868

x=3.344e^-6

%ionized = 0.34e-6/0.0868 * 100 = 0.00385%

### Question 39

Calculate the percent ionized in 9.33×10-6 M phenol–a very dilute solution. ### Explanation

Ka= 1.288e^-10=x^2/9.33e^-6

x=3.5e^-8

3.5e^-8/(9.33e^6-6) * 100 = 0.372%

### Question 40

Based on the last two problems, as a weak acid is diluted the percent ionized

Select one: ### Question 41

A solution is made by dissolving 31.4 g of NaOH in approximately 450 mL of water in a volumetric flask. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to total 500 mL of solution. Calculate the pH of the final solution. Report pH to 2 decimal places. ### Explanation

3.14g * 1 mol/ 39.997 g = 0.785 mol/0.5L = 1.57M [NaOH]

pOH= -log[OH-]

=-log(1.57)

=-0.1959

pOH + pH = 14

14 + 0.1959 = 14.1959

### Question 42

Why wait for the solution to return to room temperature before adding the rest of the water? (More than one answer is required.)

Select one or more:  ### Question 43

sodium hydroxide solution at an elevated temperature of 50ºC.

(Recall, Kw is for water’s autodissociation, that is, H2O ↔ H+ + OH-. Remember normally pH is measured at 25ºC. (Hint, this reaction is the reverse of adding a base to an acid. In other words is it endothermic or exothermic?)

Select one or more:  ### Explanation

LeChatelier’s principle is useful in answering this question.