Question

A 450-mL solution of Sr(OH)₂ has a pH of 11.28. What mass, in milligrams, of strontium hydroxide is present in this solution? Please express your numeric answer to two significant figures.

Answer 1

use:

pH = -log [H3O+]

11.28 = -log [H3O+]

[H3O+] = 5.248*10^-12 M

use:

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(5.248*10^-12)

[OH-] = 1.905*10^-3 M

Sr(OH)2 —> Sr2+ + 2OH-

So,

[Sr(OH)2] = [OH-]/2 = (1.905*10^-3)/2 = 9.525*10^-4 M

volume , V = 450 mL

= 0.450 L

number of mol,

n = Molarity * Volume

= 0.000953*0.45

= 4.286*10^-4 mol

Molar mass of Sr(OH)2,

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

= 1*87.62 + 2*16.0 + 2*1.008

= 121.636 g/mol

mass of Sr(OH)2,

m = number of mol * molar mass

= 4.286*10^-4 mol * 121.636 g/mol

= 5.2*10^-2 g

Answer: 5.2*10^-2 g