Question

A chemist dissolves 222.mg of pure potassium hydroxide in enough water to make up 80.mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25°C.) Round your answer to 2 significant decimal places.

Answer 1

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

mass(KOH)= 222 mg

= 0.222 g

use:

number of mol of KOH,

n = mass of KOH/molar mass of KOH

=(0.222 g)/(56.11 g/mol)

= 3.957*10^-3 mol

volume , V = 80 mL

= 8*10^-2 L

use:

Molarity,

M = number of mol / volume in L

= 3.957*10^-3/8*10^-2

= 4.946*10^-2 M

So,

[OH-] = 4.946*10^-2 M

use:

pOH = -log [OH-]

= -log (4.946*10^-2)

= 1.3057

use:

PH = 14 - pOH

= 14 - 1.3057

= 12.6943

Answer: 12.69