Question

You prepare a solution of Ca(OH)2 by dissolving 4.23 g of the substance in 455 mL of water. What is the pH of the solution? what is the pOH?

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)= 4.23 g

number of mol of Ca(OH)2,

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

=(4.23 g)/(74.096 g/mol)

= 5.709*10^-2 mol

volume , V = 455 mL

= 0.455 L

Molarity,

M = number of mol / volume in L

= 5.709*10^-2/0.455

= 0.1255 M

This is concentration of Ca(OH)2

So,

[OH-] = 2*[Ca(OH)2] = 2*0.1255 M = 0.251 M

use:

pOH = -log [OH-]

= -log (0.251)

= 0.600

use:

PH = 14 - pOH

= 14 - 0.600

= 13.4

Answer:

pH = 13.4

pOH = 0.600