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
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.
To calculate the final concentration, you need to use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case, you have 175 mL of a 0.45 M sodium sulfite solution and you add it to 650 mL of water. Since water does not contribute to the concentration, the final volume will simply be the sum of the volumes of the solution and water, which is 175 mL + 650 mL = 825 mL.
Now let's plug the values into the formula and solve for C2:
(0.45 M)(175 mL) = C2(825 mL)
Multiply the numbers:
78.75 = 825C2
Divide both sides of the equation by 825 to isolate C2:
78.75/825 = C2
Simplify the fraction:
0.095 = C2
Therefore, the final concentration is 0.095 M (rounded to three decimal places).