Question

What is the pH of a solution prepared by dissolving 1.4 g of Ca(OH)2 in water to make 930. mL of solution?

Express your answer using two decimal places.

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)= 1.4 g

use:

number of mol of Ca(OH)2,

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

=(1.4 g)/(74.1 g/mol)

= 1.889*10^-2 mol

volume , V = 9.3*10^2 mL

= 0.93 L

use:

Molarity,

M = number of mol / volume in L

= 1.889*10^-2/0.93

= 2.032*10^-2 M

So,

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

= 2*2.032*10^-2 M

= 4.064*10^-2 M

use:

pOH = -log [OH-]

= -log (4.064*10^-2)

= 1.391

use:

PH = 14 - pOH

= 14 - 1.391

= 12.609

Answer: 12.61