Question

The ionic compound, Mg(HCO3)2, will fully dissolve in water. If 16.03 g is dissolved in 456.0 mL water, find the following ion concentrations. The concentration of the Mg(2+) is: _____ M. The concentration of the HCO3(-) is: ______ M.

Answer 1

Molar mass of Mg(HCO3)2,

MM = 1*MM(Mg) + 2*MM(H) + 2*MM(C) + 6*MM(O)

= 1*24.31 + 2*1.008 + 2*12.01 + 6*16.0

= 146.346 g/mol

mass(Mg(HCO3)2)= 16.03 g

use:

number of mol of Mg(HCO3)2,

n = mass of Mg(HCO3)2/molar mass of Mg(HCO3)2

=(16.03 g)/(1.463*10^2 g/mol)

= 0.1095 mol

volume , V = 4.56*10^2 mL

= 0.456 L

use:

Molarity,

M = number of mol / volume in L

= 0.1095/0.456

= 0.2402 M

So,

[Mg(HCO3)2] = 0.2402 M

1 Mg(HCO3)2 gives 1 Mg2+ and 2 HCO3-

So,

[Mg2+] = [Mg(HCO3)2] = 0.2402 M

[HCO3-] = 2*[Mg(HCO3)2] = 2*0.2402 M = 0.4804 M

[Mg2+] = 0.240 M

[HCO3-] = 0.480 M