CHEM126 – CHAPTER 16 – REGULAR HOMEWORK / ACIDS I

 

CHAPTER 16 REGULAR HOMEWORK:

ACIDS I

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Question 1

Calculate the pH of a solution made by adding 34 g of sodium acetate, NaCH3COO, to 26 g of acetic acid, CH3COOH, and dissolving in water to make 600. mL of solution. Hint given in feedback.

The Ka for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places.

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Explanation

What kind of solution does this make? . . . It is a buffer solution!

CH3COOH <——>   H+    + CH3COO-

26g                                            34g

0.043297 mol        0              0.41448 mol

-x                            +x                 +x

(0.043297/0.6L)-x     x                  (0.41448/0.6L)+ x

 

Ka = 1.8e-5  = (x)(0.6908 + x)/ (0.7216 -x)      [ignore the x as negligible]

x= 0.0000188

pH= -log[H+]= -log[0.0000188]=4.72584

Question 2

Calculate the pH of a solution made by adding 80. g of sodium acetate, NaCH3COO, to 39 g of acetic acid, CH3COOH, and dissolving in water to make 500. mL of solution.

 

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Explanation

 

CH3COOH <——>   H+    + CH3COO-

39g                                            80g

0.65 mol              0              0.9753 mol

-x                            +x                 +x

(0.65/0.6L) -x         x                  (0.9753/0.6L) + x

Ka = 1.8e-5  = (x)(1.95 + x)/ (1.3 -x)      [ignore the x as negligible]

x= 0.000011997

pH= -log[H+]= 4.92

 

 

 

Question 3

Calculate the pH of a solution made by adding 33.0 g of sodium formate, NaHCOO, to 100. mL of 0.48 M formic acid, HCOOH. Hint what kind of solution is made?

The Ka for HCOOH is 1.8 x 10-4 M. As usual, report pH to 2 decimal places.

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Explanation

Buffer solution!

HCOOH       <——————->    H+        +  HCOO-

0.48*0.1                                                          33g

0.048 mol*0.1L                                            0.485 mol/0.1L

0.48                                             0                  4.85

-x                                                   +x                +x

0.48 -x                                         x                      4.85+x

 

Ka=1.8e-5=x(4.85+x)/(0.48-x)    [x is negligible]

x=0.0000178

pH=-log[H+]=4.75

Question 4

Calculate the pH of a solution made by adding 28.0 g of sodium formate, NaHCOO, to 300. mL of 0.32 M formic acid, HCOOH.

 

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Explanation

Buffer solution!

HCOOH       <——————->    H+        +  HCOO-

0.32 M                                                                 28g

0.32M                                           0                     1.37 M

-x                                                  +x                        + x

0.32-x                                          x                    1.37+x

 

 

Ka=1.8e-5=x(1.37+x)/(0.32-x)    [x is negligible]

x=0.000042044

pH=-log[H+]=4.38

 

 

Question 5

What is the pH of pure water at 25ºC?

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Question 6

When a small amount of acid is added to a non-buffered solution, there is a large change in pH.

Calculate the pH when 26.8 mL of 0.0047 M HCl is added to 100.0 mL of pure water. Comment and hint in the general feedback.

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Explanation

Note, less moles of acid are added in this example than in the following buffered example. Yet, you will see that the pH change in the buffered solution is smaller. (Hint, remember to use the total volume in your pH calculation.)

 

0.0268 L * 0.0047M = 0.00012596 mols HCl/0.1268 L = 0.00099 M

pH= -log(0.00099)= 3.003

Question 7

A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared. Calculate the initial pH of this solution.

The Ka for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places.

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Explanation

CH3COOH <——-> H+    +                   CH3COO-

0.2 M                        0                                0.1 M

-x                              +x                                 +x

0.2-x                       x                                   0.1 + x

Ka= 1.8e-5   = (x)(0.1)/(0.2)

x=0.000036

pH=-log(x) = 4.44

Question 8

A buffered solution resists a change in pH.

Calculate the pH when 25.3 mL of 0.018 M HCl is added to 100.0 mL of the above buffer.

 

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Explanation

CH3COOH <——-> H+    +                   CH3COO-

(0.199964*0.1)     (0.0253*0.018)         (0.1 + 0.000036)(0.1)

+0.0004554            -0.0004554                  -.0004554

0.02045                         0                               0.0095482

1.8e-5=x(0.0095482)/(0.02045)

x=0.00003855

pH=4.41

Question 9

Neutralization examples

Calculate how many mL of 0.100 M NaOH are needed to neutralize completely 55.0 mL of 0.0200 M HCl.

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Explanation

(0.055L) (0.02 M) = 0.0011 mols HCl

0.0011 mol NaOH/ x L = 0.1 M

x=0.011 L = 11 mL

Question 10

Calculate how many mL of 0.100 M NaOH are needed to neutralize completely 93.0 mL of 0.0400 M H2SO4 (forming Na2SO4 and water).

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Explanation

2NaOH + H2SO4 ——-> Na2SO4   + 2H20

(0.093 L)(0.04 M)  = (0.00372 mol H2SO4)

2(0.00372) mol / x L = 0.1 M NaOH

x= 0.0744 L = 74.4 mL

Question 11

Calculate how many mL of 0.100 M NaOH are needed to neutralize completely 50.0 mL of 0.0400 M acetic acid.

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Explanation

NaOH       + CH3COOH    ——->   NaCH3COO-    + H20

 

(0.05L)(0.04M) = 0.002 mols CH3COOH

0.02 mol / x  L = 0.1 M NaOH

x=0.02 L= 20 mL

Question 12

Titration of Strong Acid

Calculate the pH during the titration of 100.0 mL of 0.200 M HCl with 0.400 M NaOH. First what is the initial pH (before any NaOH is added)?

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Explanation

pH= -log(0.2) = 0.69897

Question 13

What is the pH after 28.0 mL of NaOH are added?

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Explanation

(0.1 L)(0.2 M) = (0.02 mols HCl)

(0.028 L)(0.4 M) = (0.0112 mols NaOH)

0.02-0.0112= 0.0088 mols HCl/ 0.128 L = 0.06875

pH= -log (0.06875) = 1.16273

Question 14

What is the pH after 50 mL of NaOH are added?

Explanation

(0.1 L)(0.2 M) = 0.002 mols HCl

(0.05 L)(0.4) = 0.002 mols NaOH

mols NaOH = mols HCl

pH=7

Question 15

What is the pH after 69.0 mL of NaOH are added?

 

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Explanation

(0.1)(0.2)= (0.02) mols HCl

(0.069)(0.4 M) = (0.0276 mols NaOH)

0.0276 – 0.02 = 0.0076 mols NaOH

pOH= -log (0.0076/0.169) = 1.347

pH = 14 – 1.347 = 12.65

Question 16

pH polyprotic acid

Calculate the pH of 0.033 M phosphoric acid (H3PO4, a triprotic acid).

Ka1 = 7.5 x 10-3, Ka2 = 6.2 x 10-8, and Ka3 = 4.8 x 10-13.

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Explanation

Hint, if you are doing much work, you are making the problem harder than it needs to be.

 

H3PO4 <——> H+   + H2PO4-

0.033                0               0

-x                      +x              +x

0.033-x            x                  x

 

7.5e-3 = x^2/ (0.033-x)

0.0002475 – 7.5e-3x = x^2

x=0.0124

pH= -log(0.0124) = 1.906

Question 17

Phosphoric acid is a triprotic acid:

H3PO4(aq) +H2O(l) ↔ H3O+(aq) + H2PO4-(aq) Ka1 = 7.5 x 10-3
H2PO4-(aq) +H2O(l) ↔ H3O+(aq) + HPO42-(aq) Ka2 = 6.2 x 10-8
HPO42-(aq) +H2O(l) ↔ H3O+(aq) + PO43-(aq) Ka3 = 4.8 x 10-13

Determine whether sodium monohydrogen phosphate (Na2HPO4) is neutral, basic, or acidic.

First, what is its Ka when it acts as an acid?

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Question 18

Second, what is its Kb when it acts as a base?

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Explanation

Kb= Kw/Ka= 1e-14/ 6.2e-8 = 1.613e-7

Question 19

Finally, indicate whether the HPO42- ion is neutral, basic, or acidic in solution.

 

Select one:

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Explanation

Question 20

Determine whether potassium hydrogen tartrate (KHC4H4O6) is neutral, basic, or acidic.

First, what is its Ka when it acts as an acid?

The following are for the diprotic acid, H2C4H4O6:
Ka1 = 1.0 x 10-3 and Ka2 = 4.6 x 10-5.

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Question 21

Second, what is its Kb when it acts as a base?

 

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Explanation

Kb = Kw/Ka = 1e-14/1e-3 = 1e-11

Question 22

Finally, indicate whether the HC4H4O6- ion is neutral, basic, or acidic in solution.

 

Select one:



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