(10 points) You have a standardized calcium ion (Ca ppm by mass e 10.0(3) g Ca/

Question

(10 points) You have a standardized calcium ion (Ca ppm by mass e 10.0(3) g Ca/ 1x106 g solution) from which you are making a dilution with. The standardized solution has a density of 1.03(4) mL. You transfer 10 mL of the standardized solution to a 100 mL volumetric flask with an uncalibrated (systematic uncertainty) 1 mL pipette specified to deliver 1.005) mL Yor volumetric flask is also uncalibrated and has a specified volume of 100.0(4) mL, which you fill perfectly to the line with pure H2O after your addition of the calcium solution Recall that you must also include the uncertainty in the molar mass of calcium as well (molar mass = 40.078(4) g mol). Determine the molarity including the absolute uncertainty in the final solution 4, solution which has been certified to be 10.0(3)

Explanation / Answer

Initial concentration of the solution = 10.0(3)ppm = 0.01003g/L

For the calcium solution, 40.0784g/L = 1 M

So, 0.01003g/L = x M

Use the formula M1V1 = M2V2

x = (1 X 0.01003)/ 40.0784 g/L = 2.5026 X 10-4M

Hence the molarity of the stock solution is 2.5026 X 10-4M.Now we are diluting 10ml into 100ml

Considering the calibration error of pipette and the volumetric flask,the 10.05ml of 2.5026 X 10-4M solution is diluted to 100.04ml water.

Again applying the equation, M1V1 = M2V2

molarity of the final solution=(2.5026 X 10-4M X 10.05ml)/100.04ml= 2.51414 X 10-5 M

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