Question

Calculate the molar concentration of OH negative in water solutions with the following H3O positive molar concentrations:

a) 0.044

b) 1.3 * 10 to the negative 4 power

c) 0.0087

d) 7.9 * 10 to the negative 10 power

e) 3.3 * 10 to the negative 2 power

(Please show clear handwriting and step by step) Thank you!!

Answer 1

a)

use:

[OH-] = Kw/[H3O+]

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

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(4.4*10^-2)

[OH-] = 2.3*10^-13 M

Answer: 2.3*10^-13 M

b)

use:

[OH-] = Kw/[H3O+]

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

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(1.3*10^-4)

[OH-] = 7.7*10^-11 M

Answer: 7.7*10^-11 M

c)

use:

[OH-] = Kw/[H3O+]

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

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(8.7*10^-3)

[OH-] = 1.1*10^-12 M

Answer: 1.1*10^-12 M

d)

use:

[OH-] = Kw/[H3O+]

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

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(7.9*10^-10)

[OH-] = 1.3*10^-5 M

Answer: 1.3*10^-5 M

e)

use:

[OH-] = Kw/[H3O+]

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

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(3.3*10^-2)

[OH-] = 3.0*10^-13 M

Answer: 3.0*10^-13 M