Question

-/0.1 points 24. 0/4 Submissions Used Complete the following table by calculating the missing entries. In each case indicate whether the solution is acidic or basic [H [ОН 1] pOH [OH ] acidic or basic? pH acidic 9.53 basic M M acidic 11.18 basic M acidic 5.70 x 10-10 M basic M acidic 6.40 x 10-3 M basic M Σ Σ

Answer 1

1) for row 1:

POH = 14 - pH

= 14 - 9.53

= 4.47

use:

pH = -log [H+]

9.53 = -log [H+]

[H+] = 2.951*10^-10 M

use:

pOH = -log [OH-]

4.47 = -log [OH-]

[OH-] = 3.388*10^-5 M

Since pH > pOH, this is basic in nature

Answers:

pH = 9.53

pOH = 4.47

[H+] = 2.95*10^-10

[OH-] = 3.39*10^-5

basic

2) for row 2:

use:

PH = 14 - pOH

= 14 - 11.18

= 2.82

use:

pH = -log [H+]

2.82 = -log [H+]

[H+] = 1.514*10^-3 M

use:

pOH = -log [OH-]

11.18 = -log [OH-]

[OH-] = 6.607*10^-12 M

Since pH < pOH, this is acidic in nature

Answers:

pH = 2.82

pOH = 11.18

[H+] = 1.51*10^-3

[OH-] = 6.61*10^-12

acidic

3) for row 3:

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)/(5.7*10^-10)

[OH-] = 1.754*10^-5 M

use:

pH = -log [H+]

= -log (5.7*10^-10)

= 9.2441

use:

pOH = -log [OH-]

= -log (1.754*10^-5)

= 4.7559

Since pH > pOH, this is basic in nature

Answers:

pH = 9.24

pOH = 4.76

[H+] = 5.70*10^-10

[OH-] = 1.75*10^-5

basic

4) for row 4:

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)/6.4*10^-3

[H+] = 1.562*10^-12 M

use:

pH = -log [H+]

= -log (1.562*10^-12)

= 11.8062

use:

pOH = -log [OH-]

= -log (6.4*10^-3)

= 2.1938

Since pH > pOH, this is basic in nature

Answers:

pH = 11.81

pOH = 2.19

[H+] = 1.56*10^-12

[OH-] = 6.40*10^-3

basic