Question

Complete this table of values for four aqueous solutions at 25°C. OH] POH Given: Number Number Number Solution A: 7.5 X 10 - 10 M Number Given: Number Number Solution B: 8.5x 10-6 Number Number Given: Number 10.29 Solution C: Given: Number Number Number 6.81 Solution D:

Answer 1

1) for row 1:

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(7.5*10^-10)

[OH-] = 1.333*10^-5 M

use:

pH = -log [H+]

= -log (7.5*10^-10)

= 9.1249

use:

pOH = -log [OH-]

= -log (1.333*10^-5)

= 4.8751

Answers:

[H+] = 7.5*10^-10

[OH-] = 1.3*10^-5

pH = 9.12

pOH = 4.88

2) for row 2:

use:

[H+] = Kw/[OH-]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[H+] = (1.0*10^-14)/[OH-]

[H+] = (1.0*10^-14)/8.5*10^-6

[H+] = 1.176*10^-9 M

use:

pH = -log [H+]

= -log (1.176*10^-9)

= 8.9294

use:

pOH = -log [OH-]

= -log (8.5*10^-6)

= 5.0706

Answers:

[H+] = 1.2*10^-9

[OH-] = 8.5*10^-6

pH = 8.93

pOH = 5.07

3) for row 3:

POH = 14 - pH

= 14 - 10.29

= 3.71

use:

pH = -log [H+]

10.29 = -log [H+]

[H+] = 5.129*10^-11 M

use:

pOH = -log [OH-]

3.71 = -log [OH-]

[OH-] = 1.95*10^-4 M

Answers:

[H+] = 5.1*10^-11

[OH-] = 2.0*10^-4

pH = 10.29

pOH = 3.71

4) for row 4:

use:

PH = 14 - pOH

= 14 - 6.81

= 7.19

use:

pH = -log [H+]

7.19 = -log [H+]

[H+] = 6.457*10^-8 M

use:

pOH = -log [OH-]

6.81 = -log [OH-]

[OH-] = 1.549*10^-7 M

Answers:

[H+] = 6.5*10^-8

[OH-] = 1.5*10^-7

pH = 7.19

pOH = 6.81