1)Finish the following reactions when acids are added to water.

a. H2SO4 + H2O -> HSO4- + H3O+  (remove an H+ from the acid and add to the                   base)

b. HCl + OH- ->

c. NH3 + H2O ->

d. HCl + NH3 ->

 

2) What are the conjugate bases of these acids? (Remove an H+ and write what is left)

Original acid                         Conjugate base                                              

 HNO3                                NO3- 

 H2O

  H3O+       

H2SO4

HBr

HCO3-

 

3) What are the conjugate acids of these bases? (add an H+)

Original Base                        Conjugate Acid                                             

OH-                                                            H2O

H2O      

HCO3-

SO42-

ClO4-

 

4) Which of the following represent conjugate acid-base pairs? 

1. H2O, H3O+
2. OH-, HNO3
3. H2SO4, SO4-2
4. HC2H3O2, C2H3O2-

 

5) Calculate the [H+] in a solution in which [OH-] = 2.0 x 10-2 M. (I used: [H+][OH-]=1x10-14)

Is this solution acidic, neutral, or basic? 

 

 6) Calculate the [OH-] in a solution in which [H+] = 3.99 x 10-5 M. (I used: [H+][OH-]=1x10-14)

Is this solution acidic, neutral, or basic?

 

7) Convert each of the following into pH. (-log[H+] or -log[OH-] and then use pH + pOH = 14 if necessary)

Identify each as an acidic(A), neutral(N), or basic(B). 

a. 0.0015 M H+

b. 5.0 x 10-9 M H+

c. 3.27 x 10-4 M OH-

d. 1.00 x 10-12 M OH-

 

8) Convert each of the following into pOH. 

Identify each as an acidic(A), neutral(N), or basic(B).

a. 0.783 M OH-

b. 6.87 x 10-12 M OH-

c. 1.1 x 10-9 M H+

d.  0.0032 M H+

 

9) What is the pH, pOH, [H+], or [OH-] for a [0.156] molar solution of hydrochloric acid (releases H+)? 

 

10) What is the pH, pOH, [H+], or [OH-] for a 3.2 molar solution of sodium hydroxide (releases OH-)? 

Ans 1) a)

H2SO4 + 2H2O —> 2H3O+ + SO4²-

b) HCl + OH- —> Cl- + H2O

c) NH3 + H2O —> NH4OH

d) HCl + NH3 —> NH4Cl

 

Ans 2) Conjugate base is formed when acid loses H+

HNO3 —> NO3-

H2O —> OH-

H3O+ —> H2O

H2SO4 —> HSO4-

HBr —> Br-

HCO3- —> CO3²-

 

Ans 3) Conjugate acid is formed when H+ is added to base

OH- —> H2O

H2O —> H3O+

HCO3- —> H2CO3

SO4²- —> HSO4-

ClO4- —> HClO4

 

Ans 4) Conjugate acid-base pairs are represented by

a) H2O, H3O+

and d) HC2H3O2 , C2H3O2-

 

It can be explained by the reaction;

HC2H3O2 + H2O —> C2H3O2- + H3O+

where HC2H3O2 = Acid

H2O = Base

C2H3O2- = Conjugate base

H3O+ = Conjugate acid

 

Ans 5) We know that,

Kw = [H+] [OH-] = 10^-14

[OH-] = 2.0 × 10^ -2

[H+] = 10^-14 / 2.0 × 10^ -2

[H+] = 0.5 × 10^-12

pH = -log [H+] = -log ( 0.5 × 10^-12)

pH = 12.30

The solution is basic

 

Ans 6)

[H+] = 3.99 × 10^-5

[OH-] = 10^-14 / 3.99 × 10^-5

[OH-] = 0.25 × 10^ -9

pH = -log(3.99 × 10^-5 )

pH = 4.39

The solution is acidic.

 

Ans 7) a) pH = -log [H+ ]

pH = - log (0.0015)

pH = 2.82

The solution is acidic.

b) pH = -log ( 5× 10^-9)

pH = 8.30

The solution is basic.

c) [OH-] = 3.27 × 10 ^-4

[H+] = 10^-14 / 3.27 × 10 ^-4

[H+] = 0.305 × 10^-10

pH = -log(0.305 × 10^-10)

pH = 10.51

The solution is basic.

d) [OH-] = 1× 10^-12

[H+] = 10^-14/ 1×10^-12 = 10^-2

pH = 2

The solution is acidic.

 

Ans 8) a) [OH-] = 0.783

pOH = -log [OH-]

pOH = -log ( 0.783)

pOH = 0.11

pH = 14- pOH = 14 - 0.11 = 13.89

The solution is basic.

b) [OH-] = 6.87 × 10^ -12

pOH = -log (6.87 × 10^ -12)

pOH = 11.15

pH = 14- 11.15 = 2.85

The solution is acidic.

c) [H+] = 1.1 × 10^-9

pH = -log (1.1 × 10^-9) = 8.95

pOH = 14-8.95 = 5.05

The solution is basic.

d) [H+] = 0.0032

pH = -log ( 0.0032) = 2.49

pOH = 14-2.49 = 11.51

The solution is acidic.

 

Ans 9)

Given ; HCl = 0.156 M

[H+] = 0.156 M

pH = -log (0.156) = 0.80

[OH-] = 10^-14/ 0.156 = 6.41 × 10^-14

pOH = 14 - pH = 14-0.80 = 13.2

 

Ans 10) 3.2 M NaOH

[OH-] = 3.2

pOH = -log [OH-] = -0.505

[H+] = 10^-14 / 3.2 = 0.312 × 10^ -14

pH = -log (0.312 × 10^ -14 ) = 14.50

(If concentration of hydroxide ions exceeds one molar then pH can go higher than 14)