Yesterday’s Homework

Page 611 # 19

Page 612 # 20

Calculate the pH of solutions having the following ion concentrations at 298o K.

[H+] = 1.0 x 10-2

log [H+] = -2

pH = 2

Calculate the pH of solutions having the following ion concentrations at 298o K.

[H+] = 3.0 x 10-6

log [H+] = 0.477 – 6

= -5.523

pH = 5.523

Calculate the pH of solutions having the following ion concentrations at 298o K.

[OH-] = 8.2 x 10-6

log [OH-] = 0.913 – 6

= -5.087

pOH = 5.087

Calculate the pH of solutions having the following ion concentrations at 298o K.

pOH = 5.087

pH + pOH = 14

pH = 14 – pOH

= 14 – 5.087

= 8.913

Calculate the pH of solutions having the following ion concentrations at 298o K.

[OH-] = 1.0 x 10-6

log [OH-] = -6

pOH = 6

pH = 14 – 6

= 8

Calculate the pH of solutions having the following ion concentrations at 298o K.

[OH-] = 6.5 x 10-4

log [OH-] = 0.812 - 4

pOH = 3.188

pH = 14 – 3.188

= 10.812

Calculate the pH of solutions having the following ion concentrations at 298o K.

[H+] = 3.6 x 10-9

log [H+] = 0.556 - 9

pH = 8.444

pOH = 14 – 8.444

= 5.556

[H+] = .025

Calculate the pH of solutions having the following ion concentrations at 298o K.

[H+] = 2.5 x 10-2

log [H+] = 0.397 – 2

pH = 1.603

pOH = 14 – 1.603

= 12.397

Calculate the pH of solutions having the following ion concentrations at 298o K.

Now let’s try it the other way around…

Given the pH, calculate the [H+] and [OH-] for the following solution.

pH = 7.40

-log [H+] = 7.40

log [H+] = -7.40

= .60 – 8

[H+] = 4.0 x 10-8

Given the pH, calculate the [H+] and [OH-] for the following solution.

pOH = 14 – 7.40

= 6.60

log [OH-] = -6.60

= .40 – 7

[OH-] = 2.5 x 10-7

One last thing…If you have a solution of 1.0 M HCl, what is the [H+]?

One last thing…HCl is a strong acid.

That means it dissociates completely.

If [HCl] = 1.0, then [H+] = 1.0

One last thing…Same with strong bases.

If [NaOH] = 1.0, then [OH-] = 1.0

Homework

Page 614

Problems 21 & 22