Question

2. (2 pts) Find pH of solution obtained by dissolving 74 mg of calcium hydroxide in 100.0 mL of water.

Answer 1

Molar mass of Ca(OH)2,

MM = 1*MM(Ca) + 2*MM(O) + 2*MM(H)

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

mass(Ca(OH)2)= 0.074 g

use:

number of mol of Ca(OH)2,

n = mass of Ca(OH)2/molar mass of Ca(OH)2

=(7.4*10^-2 g)/(74.1 g/mol)

= 9.987*10^-4 mol

volume , V = 1*10^2 mL

= 0.1 L

use:

Molarity,

M = number of mol / volume in L

= 9.987*10^-4/0.1

= 9.987*10^-3 M

So,

[OH-] = 2*[Ca(OH)2]

= 2*9.987*10^-3 M

= 1.997*10^-2 M

use:

pOH = -log [OH-]

= -log (1.997*10^-2)

= 1.6996

use:

PH = 14 - pOH

= 14 - 1.6996

= 12.3004

Answer: 12.30