You have 75.0 mL of a 2.50 M solution of Na2CrO4 (aq). You also have 125 mL of a 1.79 M solution of AgNO3 (aq). Calculate the concentration of CrO42 - after the two solutions are mixed together.

Question

Chemistry

Alisson Mcbride

9 September, 21:39

You have 75.0 mL of a 2.50 M solution of Na2CrO4 (aq). You also have 125 mL of a 1.79 M solution of AgNO3 (aq). Calculate the concentration of CrO42 - after the two solutions are mixed together.

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Anne Blackburn · Answer 1

The reaction of Na2CrO4 (aq) and AgNO3 (aq) is as follows:

2AgNO3 (aq) + Na2CrO4 (aq) ⇒Ag2CrO4 (s) + 2NaNO3 (aq).

During this reaction, part of the CrO42 - reacts with Ag + and precipitates out of the solution, part of the CrO42 - (excess amount) remains in the solution.

To find out how much CrO42 - is reacted:

Moles of initial CrO42 - = 0.075 L * 2.5 M = 0.1875 mole

Moles of initial Ag + = 0.125 L * 1.79 M = 0.2238 mole

The reaction ratio between CrO42 - and Ag + is 1:2 according to the equation. So moles of CrO42 - that is reacted is 0.2238 mole/2 = 0.1119 mole. Therefore, moles of CrO42 - that remains in the solution is 0.1875mole-0.1119mole = 0.0756 mole

So the final concentration of CrO42 - in the solution is 0.0756mole / (0.075L+0.125L) = 0.378 M.