Question

What mass of KOH is necessary to prepare 867.0 mL of a solution having a pH = 11.44?

Answer 1

Given:

pH = 11.44

use:

pH = -log [H+]

11.44 = -log [H+]

[H+] = 3.631*10^-12 M

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)/(3.631*10^-12)

[OH-] = 2.754*10^-3 M

So,

[KOH] = 2.754*10^-3 M

volume , V = 8.67*10^2 mL

= 0.867 L

use:

number of mol,

n = Molarity * Volume

= 2.754*10^-3*0.867

= 2.388*10^-3 mol

Molar mass of KOH,

MM = 1*MM(K) + 1*MM(O) + 1*MM(H)

= 1*39.1 + 1*16.0 + 1*1.008

= 56.108 g/mol

use:

mass of KOH,

m = number of mol * molar mass

= 2.388*10^-3 mol * 56.11 g/mol

= 0.134 g

Answer: 0.134 g