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CHAPTER 16 REGULAR HOMEWORK:

ACID II

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Question 1

Titration of a weak acid

Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH. First what is the initial pH (before any NaOH is added)?

The Ka for HOCl is 3.0 x 10-8 M.

Need help? Hint, given in general feedback.

Correct


 

Explanation

 Make ICE table and use simplified method of successive approximations (no iterations needed).


HClO     -------->    H+      +          ClO-

0.4                         0                     0

-x                            +x                    +x

0.4-x                        x                    x


Ka= 3e-8= x^2/(0.4-x)                 x is negligible

x=0.000109545

pH=-log(x) = 3.96

 

Question 2

How many mL of NaOH are added to reach the equivalence point?

Correct

 

Explanation

 (0.4 M) (0.01 L)  = 0.004 mol HClO

0.004 mol NaOH / x L  = 0.5M NaOH

 x = 8 mL

 

Question 3

What is the pH after 1.80 mL of NaOH are added?

Hint given in general feedback.

Correct


 

Explanation

 Hint, note that the amount NaOH added is less than what is needed to completely neutralize acid. So, you have made a buffer. Use buffer equation.


(0.0018L)(0.5)=0.0009 mols NaOH

0.004 mols HClO

HClO - NaOH =  0.004-0.0009= 0.00310 mol HClO

0.00310 mol HClO/ (0.01 + 0.0018)L  = 0.2627 M HClO

pH= 7.5229 + log(0.0009/ (0.01+0.0018)/0.2627 M = 6.98578

 

Question 4

What is the pH after 5.60 mL of NaOH are added?

Hint given in general feedback.

Correct


 

Explanation

 Hint, note that the amount NaOH added is less than what is needed to completely neutralize acid. So, you have made a buffer. Use buffer equation.


(0.5 M)(0.0056 L) = 0.0028 mols NaOH

                                   -0.004 HClO

                                    0.0012 mols HClO

pH= 7.5229 + log (0.0028/ x L)/ (0.0012/ x L)

pH=7.89

 

Question 5

What is the pH at the equivalence point?

Notice that the pH of a weak acid at the equivalence point is basic.

Correct

 

Explanation

 ClO-  + H20    <----->     HClO    +     OH-

[ClO-]   =   0.004 mols /0.018 L   = 0.222 M

ClO-       <--------->     OH-      + HClO

0.222                           0              0

-x                                +x               +x

0.222-x                     x                 x


Kb= Kw/Ka= 1e-14/3e-8=3.33e-7

Kb= [HClO][OH-]/[ClO-] = x^2/0.222-x

x=2.72e-4

pOH=-log(x)= 3.588

pH= 14- 3.588=10.43

 

Question 6

What is the pH after 8.80 mL of NaOH are added?

Hint given in general feedback.

Correct

 

Explanation

 Hint, note that the amount NaOH added is more than what is needed to completely neutralize acid. So, you have strong base plus a weak base.


(0.0088L)(0.5 M) = 0.0044 mols OH-

 [OH-]= (0.0044 mol - 0.004 HClO)/ 0.0188L = 0.021276596

pOH= -log(OH-)= 1.672

pH=12.33

 

Question 7

What is the pH when half the acid has been neutralized?

Hint given in general feedback.

Correct

 

Explanation

 This is the easiest problem. Think buffer eq. . .

pH=pKa

=-log(3e-8)

=7.52288

 

Question 8

What is the pH after 11.60 mL of NaOH are added?

Hint given in general feedback.
Correct

 

Explanation

 Hint, note that the amount NaOH added is more than what is needed to completely neutralize acid. So, you have strong base plus a weak base.


(0.0116 L)(0.5M)=0.0058 mols NaOH

[OH-]= (0.0058 mol - 0.004 mol HClO)/ 0.0216 L = 0.0833

pH= 14 - (-log(0.0833))= 12.92

 

Question 9

Chloropropionic acid, ClCH2CH2COOH is a weak monoprotic acid with Ka = 7.94 x 10-5 M. Calculate the pH at the equivalence point in a titration 35.8 mL of 0.93 M chloropropionic acid with 0.1 M KOH.

Correct

 

Explanation

 (0.0358 L)(0.93) = 0.033294 mols 

0.033294 mol/ x L = 0.1 KOH

x=0.33294


ClCH2CH3COO-                        ------>   OH-     + ClCH2CH2COOH

0.33294 mol/(0.33294+0.0358L)          0                   0


Kb= Kw/Ka = 1e-14/7.94e-5= 1.259e-10

1.259e-10 = x^2/(0.09029-x)                 x is negligible on the bottom

x=0.00000337

pOH=-log(x)=5.47

pH=8.53

 

Question 10

Choose the best indicator from the table below for the above titration.


Select one:


 
 
 
  Correct

 

 

 

Question 11

Indicators

Phenolphthalein is a commonly used indicator that is colorless in the acidic form (pH < 8.3) and pink in the basic form (pH > 10.0). It is a weak acid with a pKa of 8.7. What is the ratio of the conjugate base concentration to the acid concentration, that is, [B- ]/[HB] when the indicator switches to the acidic color?


Correct

 

Explanation

 Recall, an indicator is a buffer, that is, use the buffer equation.


[base]/[acid]=10^(pH-pKa)


8.3 = 8.7 + logA

logA= -0.4

10^logA=10^-0.4

A=0.398

 

Question 12

Use the result to the above question to determine what percent is in the acidic form when the acid color is first apparent?

Correct

 

Explanation

 0.398/1     base/acid

% acid = acid / acid+base= 1/1.398  = 71.53%

 

Question 13

What is [B- ]/[HB] when the indicator switches to the basic color?

Correct

 

Explanation

 10 = 8.7 + log A

[base]/[acid] = 10^(10-8.7) = 19.95

 

Question 14

Use the result to the above question to determine what percent is in the basic form when the base color is first apparent?

Correct

 

Explanation

 19.95/20.95=95.23%

 

Question 15

Determine the equivalent acid-base equilibrium that results when each of the following pairs of solutions is mixed.

25.0 mL of 0.50 M NaOH + 12.0 mL of 1.50 M HCl

Pick two initial conditions from first four choices. Pick one choice from remaining choices to indicate type of equivalent solution.


Select one or more:

Correct
Correct
Correct



 

Explanation

 

Question 16

20.0 mL of 0.25 M HCOOH + 10.0 mL of 0.50 M KOH


Select one or more:

Correct


Correct




Correct

 



 

Question 17

100 mL of 0.20 M CH3COOH + 50mL of 0.10 M NaOH


Select one or more:

Correct


Correct


Correct

 

 

 

Question 18

100 mL of 0.20 M CH3COOH + 50mL of 0.10 M HI


Select one or more:

Correct
Correct




Correct

 

 

 

Question 19

100 mL of 0.20 M CH3COOH + 50mL of 0.40 M NaOH


Select one or more:


Correct

 
Correct

 


Correct

 

 

 

Question 20

100 mL of 0.10 M CH3COOH + 10mL of 0.10 M NaOH


Select one or more:


Correct

 
  Correct



Correct

 

 

 

 

Question 21


100 mL of 0.20 M CH3COOH + 60mL of 0.40 M NaOH


Select one or more:

Correct

 
Correct

 



Correct

 

 

 

Question 22

100 mL of 0.10 M CH3COOH + 80mL of 0.10 M Ba(OH)2


Select one or more:


Correct


Correct


 

Correct

 

 

 

Question 23

What is the approximate pKa for the weak acid.


Correct

 

 

 

Question 24

10.0 mL of a 0.10 M weak acid is titrated with 0.10 M NaOH. Based upon the following graph, what kind of acid is it?


Select one:


Correct

 

 

 

Question 25

Very crudely, what is the pKa1 for the weak acid in previous graph?

Correct

 

 

 

Question 26


Very crudely, what is the pKa2 for the weak acid in the previous graph?

Correct

 

 

 

Question 27

1.305 g of a solid white acid are dissolved in water and completely neutralized by the addition of 30.63 mL of 0.593 M NaOH. Calculate the molar mass of the acid, assuming it to be a monoprotic acid.

Correct

 

Explanation

 0.3063L (0.593 M ) = 0.0182 mols

 MM= 1.305g/0.0182 mol = 71.85 

 

Question 28

If additional experiments indicate that the acid is diprotic, what is its molar mass?


Select one:



Correct

 

Explanation

H2SO4 + 2 NaOH  <------> Na2SO4   + H20

0.0182/2     =    0.00825

MM = 1.305/0.00825  = 158   (double 79)

    

 

Question 29

Concentrated hydrochloric acid is 38% HCl by weight and has a density of 1.19 g/mL. A solution is prepared by measuring 78 mL of the concentrated HCl, adding it to water, and diluting to 0.900 L. Calculate the approximate molarity of this solution from the volume, percent composition, and density.

Correct

 

Explanation


(The exact M could be determined from a titration using a standardized solution of a strong base.)


78 mL * 1.19 g/ mL  = 92.82 g

92.82*.38  = 35.3716 g * 1 mol/ 36.46 g  = 0.967

0.967 mol / 0.9 L = 1.07 M