Question

Concentration of calcium in a sample may be determined by precipitation using the chromate ion, Crom-.

Ca2+ (aq) + Cro - (aq) — CaCro,(s)
A chemist combined 0.180 L of an unknown calcium solution with an excess of ammounium chromate. This resulted in the
precipitation of calcium chromate. The mass of the precipitate was 282.7 mg. What was the molar concentration of Ca2+ in the
original sample?
[Ca2+] =? M

Answer 1

The molar concentration of calcium ions (Ca2+) in the original sample was calculated to be 0.01006 M, using the mass of calcium chromate precipitate.

Step-by-step explanation:

The question involves calculating the molar concentration of calcium ions (Ca2+) in an aqueous solution using a precipitation reaction with chromate ions. To determine the concentration, we used the mass of the calcium chromate precipitate formed in the reaction. The molar mass of calcium chromate (CaCrO4) is approximately 156.07 g/mol. Given that 282.7 mg (0.2827g) of CaCrO4 was formed:

Firstly, we convert the mass of the precipitate to moles:

0.2827 g / 156.07 g/mol = 1.811 x 10-3 mol

Since the precipitation reaction is 1:1, the moles of Ca2+ in the original solution are the same as the moles of CaCrO4. Hence, the concentration of Ca2+ is:

1.811 x 10-3 mol / 0.180 L = 0.01006 M