Given, we need to prepare 250 mL of 0.500 M H3PO4 solution.
First, let's calculate the required volume of the concentrated 14.6 M H3PO4 solution that we need to dilute to get the required 250 mL of 0.500 M solution.
Using the dilution formula, C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
C1 = 14.6 M (initial concentration of concentrated H3PO4)
V1 = We need to find this (volume of concentrated H3PO4 required)
C2 = 0.500 M (final concentration needed in the 250 mL flask)
V2 = 0.250 L (final volume needed, since 250 mL = 0.250 L)
Plugging in these values into the equation:
14.6 M * V1 = 0.500 M * 0.250 L
Now, solve for V1:
V1 = (0.500 M * 0.250 L) / 14.6 M
V1 = 0.125 / 14.6
V1 ≈ 0.00856 L
To convert this to milliliters, multiply by 1000:
V1 ≈ 8.56 mL
Now we have the required volume of the concentrated H3PO4 solution.
Procedure:
1. Put on proper lab safety gear, including gloves, goggles, and a lab coat.
2. Take a clean, dry 250 mL volumetric flask and a burette.
3. Using a pipette, carefully measure out approximately 8.56 mL of the 14.6 M H3PO4 from the 2.50 L stock bottle.
4. Transfer this measured volume of concentrated H3PO4 into the 250 mL volumetric flask.
5. Slowly add distilled water (deionized water) to the volumetric flask, rinsing the sides of the flask and making sure any residual H3PO4 is washed into the solution.
6. Fill the flask with distilled water up to the 250 mL mark. Mix the solution well by inverting the flask several times.
7. You now have a 250 mL flask containing a 0.500 M H3PO4 solution.
8. Label the flask with the concentration, date, and your name.
Calculations summary:
Initial concentration (C1) = 14.6 M
Final concentration (C2) = 0.500 M
Final volume (V2) = 0.250 L
Required volume of concentrated H3PO4 (V1) = 8.56 mL
You are a work study student in our chemistry department. Amy, your supervisor has just asked you to prepare 250ml of 0.500M H3PO4 solution for tomorrow's undergraduate experiment. The Stock Solutions cabinet is under the Stockroom Explorer. You will find find a 2.50 liter bottle containing 14.6M H3PO4. Please prepare a flask containing 250ml of a 0.500M solution. To ensure proper credit, please write your calculations in the space given below.
Need some help?
In designing your procedure, it is useful to remember that diluting a substance does not change the amount of that substance. So the number of moles of the substance before the dilution is the same as the number of moles after the dilution (conservation of matter). The number of moles in solution is given by the concentration (C in moles/liter) multiplied by the volume (V in liters). So the conservation of matter implies:
C1V1 = C2V2
where C is concentration and V is volume. 1 labels the solution before the dilution and 2 labels the solution after the dilution. (This expression is based on the number of moles of the substance between the same before and after the dilution, with C (moles/liter) * V (liter) = moles).
Important: Please describe your complete procedure and the key quantities you measure. Points are based on whether or not you explain your procedure in sufficient detail for us to know what you did. You are not graded on the method you used; all approaches that produce accurate results are fine.
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