Question

what is the concentration of each ion in the solution when 47.3 g of aluminum nitrate (Al(NO3)3) is dissolved in 1400 mL of water

Answer 1

1. We find the amount of substance (mol) of Al(NO₃)₃ . To do this, we divide the mass of aluminum nitrate by its molecular mass (213gr/mol):

n = 47,3 / 213 = 0,222 mol

2. Al(NO₃)₃ => Al³⁺ + 3 NO₃^–

1mol — 1mol — 3mol

0,222mol — X mol — Y mol

X = 0,222 * 1 / 1 = 0,222 mol Al³⁺

Y= 0,222 * 3 / 1 = 0,666 mol NO₃^–

3. We find the concentration of molarity of each ion. To do this, we divide the mole of the ion (n) by the volume (litres) of solution:

C_M (Al³⁺) = 0,222 mol / 1,4 l = 0,159 mol/l

C_M (NO₃^–) = 0,666 mol / 1,4 l = 0,476 mol/l

ANSWER: Al³⁺ = 0,159 mol/l

NO₃^– = 0,476 mol/l