Question

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. 0.800 mole of Ca(NO3)2 in 600.0 mL of solution MCa2+ = ______  M MNo3- = _______ M

Answer 1

1 mole of Ca(NO3)2 dissociates to give 1 mole of Ca2+ and 2 moles of NO3-

Ca(NO3)2 1Ca2+(aq) + 2(NO3)-(aq)

For Ca²⁺

1 mol Ca(NO3)2 dissociates to give 1 mole Ca²⁺
therefore 0.800 mol Ca(NO3)2 0.800 mol Ca²⁺

The solution will contain 0.800 moles Ca²⁺

Molarity = moles / Litres
Molarity Ca2+ = 0.800 moles / 0.6000 L
= 1.333 M Ca2+

for NO₃^-

1 moles Ca(NO3)2 dissociates to give 2 moles (NO3)-
0.800 moles Ca(NO3)2 (2 x 0.800) moles (NO3)-
= 1.60 moles (NO3)- present in the solution

Molarity = moles / litres
Molarity (NO3)- = 1.6 mol / 0.6000 L
= 2.666 M (NO3)-

therefore 0.800 mol Ca(NO3)2 containing 1.333 M of Ca²⁺ and 2.666 M of No₃^-