Asked by Hellen
You have 75.0 mL of a 2.50 M solution of Na_2CrO_4. You also have 125 mL of a 1.52 M solution of AgNO_3(aq). Calculate the concentration of CrO_4^2- after the two solutions are mixed together.
Answers
Answered by
DrBob222
75.0 mL of a 2.50 M solution of Na_2CrO_4. You also have 125 mL of a 1.52 M solution of AgNO_3(aq). Calculate the concentration of CrO_4^2- after the two solutions are mixed together.
millimoles Na2CrO4 = mL x M 75.0 x 2.50 = 187.5
millimoles AgNO3 = 125 x 1.52 = 190
........2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2 NaNO3
I.........190..............0........................0...............0
add....................187.5.............................................
C......-190..........-95.......................+95.............+190
E.........0..............92.5.....................95................190
M Na2CrO4 = 92.5 millimoles/200 mL = ? M.
Technically the CrO4^2- from the sparingly soluble Ag2CrO4 should be added to this but the amount from Ag2CrO4 is negligible.
millimoles Na2CrO4 = mL x M 75.0 x 2.50 = 187.5
millimoles AgNO3 = 125 x 1.52 = 190
........2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2 NaNO3
I.........190..............0........................0...............0
add....................187.5.............................................
C......-190..........-95.......................+95.............+190
E.........0..............92.5.....................95................190
M Na2CrO4 = 92.5 millimoles/200 mL = ? M.
Technically the CrO4^2- from the sparingly soluble Ag2CrO4 should be added to this but the amount from Ag2CrO4 is negligible.
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