Barium chromate, BaCrO4(s) is an insoluble yellow solid. Determine the concentration of barium ions in a solution made by mixing 50.0 mL of a 0.150 mol/L solution of barium nitrate, Ba(NO3)2(aq), with 50.0 mL of a 0.150 mol/L solution of potassium chromate, K2CrO4(aq)
1 Answer | Add Yours
First let's write out the chemical equation.
Ba(NO3)2 + K2CrO4 --> BaCrO4 + 2KNO3
When the two reagents are mixed, BaCrO4 will precipitate from solution and the KNO3 will remain dissolved in water. Now let's calculate the amount of barium and chromate present from the reactants.
0.15 mol/L * 0.05 L = 0.0075 moles Ba ions
0.15 mol/L * 0.05 L = 0.0075 moles CrO4 ions
This means that there are an equal number of moles of Ba as there are CrO4. Since the two react together in a 1:1 ratio to form the insoluble solid, this means that the 0.0075 moles of Ba will fully react with the 0.0075 moles of CrO4 to make 0.0075 moles of BaCrO4. Since the two ions are present in equal amounts and fully react, there will be no barium ions left in solution at the end of the reaction. So the answer is 0 moles/L.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes