A chemist combined 0.440 L of an unknown calcium solution with an excess of ammonium chromate. This resulted in the precipitatio

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A chemist combined 0.440 L of an unknown calcium solution with an excess of ammonium chromate. This resulted in the precipitatio

n of calcium chromate. The mass of the precipitate was 346.7 mg. What was the molar concentration of Ca2 in the original sample

Chemistry

1 answer:

Rina8888 [55] · Answer 1 · 2023-12-08T05:50:34+03:00

The concentration of the original calcium ions is 0.005 M

<h3>What is concentration?</h3>

The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.

As such we have the equation;

Ca^2+(aq) + (NH4)2CrO4(aq) --------> CaCrO4(s) + 2NH4^+(aq)

Number of moles of the precipitate = 346.7 * 10^-3 g/156 g/mol

= 0.0022 moles

Now;

1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by 0.0022 moles of CaCrO4.

Given that the volume of the solution is 0.440 L, the concentration of the solution is; 0.0022 moles/0.440 L

= 0.005 M

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