Question

3. Calculate and insert the proper values to complete the following table: (8pts) РОН [ОН-] М Acidic or Basic (H+J M PH Solution 8.21 5.79 1.78 x 10 3.70 x 10

please help

Answer 1

1) for row 1:

use:

PH = 14 - pOH

= 14 - 8.21

= 5.79

use:

pH = -log [H+]

5.79 = -log [H+]

[H+] = 1.622*10^-6 M

use:

pOH = -log [OH-]

8.21 = -log [OH-]

[OH-] = 6.166*10^-9 M

Since pH < pOH, this is acidic in nature

Answers:

pH = 5.79

pOH = 8.21

[H+] = 1.62*10^-6

[OH-] = 6.17*10^-9

acidic

2) for row 2:

POH = 14 - pH

= 14 - 5.79

= 8.21

use:

pH = -log [H+]

5.79 = -log [H+]

[H+] = 1.622*10^-6 M

use:

pOH = -log [OH-]

8.21 = -log [OH-]

[OH-] = 6.166*10^-9 M

Since pH < pOH, this is acidic in nature

Answers:

pH = 5.79

pOH = 8.21

[H+] = 1.62*10^-6

[OH-] = 6.17*10^-9

acidic

3) for row 3:

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)/1.78*10^-9

[H+] = 5.618*10^-6 M

use:

pH = -log [H+]

= -log (5.618*10^-6)

= 5.2504

use:

pOH = -log [OH-]

= -log (1.78*10^-9)

= 8.7496

Since pH < pOH, this is acidic in nature

Answers:

pH = 5.25

pOH = 8.75

[H+] = 5.62*10^-6

[OH-] = 1.78*10^-9

acidic

4) for row 4:

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)/(3.7*10^-12)

[OH-] = 2.703*10^-3 M

use:

pH = -log [H+]

= -log (3.7*10^-12)

= 11.4318

use:

pOH = -log [OH-]

= -log (2.703*10^-3)

= 2.5682

Since pH > pOH, this is basic in nature

Answers:

pH = 11.43

pOH = 2.57

[H+] = 3.70*10^-12

[OH-] = 2.70*10^-3

basic