A stock solution of potassium chromate is 1.20 M

CHEMISTRY · Multi-part · A Level · Chemistry · Amount of substance · Amount of substance

Problem

A stock solution of potassium chromate is 1.20 M. A technician transfers 10.00 mL of the stock solution to a volumetric flask and dilutes the sample to 250.0 mL with distilled water. The entire diluted solution is then poured into a beaker containing 100.0 mL of 0.650 M barium nitrate. The reaction that occurs is K2CrO4(aq)+Ba(NO3)2(aq)→BaCrO4(s)+2 KNO3(aq)\text{K}_2\text{CrO}_4(aq) + \text{Ba(NO}_3)_2(aq) \rightarrow \text{BaCrO}_4(s) + 2\,\text{KNO}_3(aq) Assume that volumes are additive and that the precipitate does not dissolve. (a) Calculate the molarity of CrO42−\text{CrO}_4^{2-} ions in the diluted potassium chromate solution before it is mixed with barium nitrate. (b) Determine the number of moles of Ba2+\text{Ba}^{2+} ions present in the barium nitrate solution. (c) Identify which ion, CrO42−\text{CrO}_4^{2-} or Ba2+\text{Ba}^{2+}, is the limiting reactant after the two solutions are mixed. Show your reasoning. (d) Calculate the mass, in grams, of barium chromate precipitate that forms. (e) After precipitation is complete, compute the concentration (molarity) of the ion that is present in excess in the final reaction mixture.

Related theory

Browse the problem library