Question

Complete the following table:

[ H3O+ ]	[ HO- ]	pH	Acidic, Basic, or Neutral
		5.0
	1 X 10 10-11 M
	8.5 X 10-4 M
3.9 X 10-3 M

Answer 1

1) for row 1:

POH = 14 - pH

= 14 - 5

= 9

we have below equation to be used:

pH = -log [H+]

5 = -log [H+]

log [H+] = -5

[H+] = 10^(-5)

[H+] = 1*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

9 = -log [OH-]

log [OH-] = -9

[OH-] = 10^(-9)

[OH-] = 1*10^-9 M

Since pH < pOH, this is acidic in nature

[H3O+] = 1*10^-5

[OH-] = 1*10^-9

pH = 5

It is acidic

2) for row 2:

we have below equation to be used:

[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)/1.0E-11

[H+] = 1*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (1*10^-3)

= 3

we have below equation to be used:

pOH = -log [OH-]

= -log (1*10^-11)

= 11

Since pH < pOH, this is acidic in nature

[H3O+] = 1*10^-3

[OH-] = 1*10^-11

pH = 3

It is acidic

3) for row 3:

we have below equation to be used:

[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)/8.5E-4

[H+] = 1.176*10^-11 M

we have below equation to be used:

pH = -log [H+]

= -log (1.176*10^-11)

= 10.9294

we have below equation to be used:

pOH = -log [OH-]

= -log (8.5*10^-4)

= 3.0706

Since pH > pOH, this is basic in nature

[H3O+] = 1.176*10^-11

[OH-] = 8.5*10^-4

pH = 10.9294

It is basic

4) for row 4:

we have below equation to be used:

[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)/(3.9*10^-3)

[OH-] = 2.564*10^-12 M

we have below equation to be used:

pH = -log [H+]

= -log (3.9*10^-3)

= 2.4089

we have below equation to be used:

pOH = -log [OH-]

= -log (2.564*10^-12)

= 11.5911

Since pH < pOH, this is acidic in nature

[H3O+] = 3.9*10^-3

[OH-] = 2.564*10^-12

pH = 2.4089

It is acidic