Asked by bri
What would be the final concentration if 175mL of a 0.45M solution of sodium sulfite Na2SO3 are added to 650mL of water?
Please reply with the steps on how to do this question so I can learn. I can't find examples or anything related in my textbooks and I take it via correspondence, so no class instructor to ask for help. Thank you
Please reply with the steps on how to do this question so I can learn. I can't find examples or anything related in my textbooks and I take it via correspondence, so no class instructor to ask for help. Thank you
Answers
Answered by
DrBob222
First, one must assume the volumes are additive; i.e., that 175 mL of the acid adds to 650 mL of water to give a total volume of 175+650 = 825.0 mL. In general volumes are not additive but the error in this assumption is small enough to be neglected.
I do these this way.
0.45M x (175/850) = ? M.
You may prefer to use the dilution formula which is
M1V1 = M2V2
0.45*125 = M2*850
Solve for M2.
Often in this kind of problem you are given the TOTAL volume of V2; remember that you don't have anything to add if the total volume is given.
I do these this way.
0.45M x (175/850) = ? M.
You may prefer to use the dilution formula which is
M1V1 = M2V2
0.45*125 = M2*850
Solve for M2.
Often in this kind of problem you are given the TOTAL volume of V2; remember that you don't have anything to add if the total volume is given.
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