You are a work study student in our chemistry department. Amy, your supervisor has just asked you to prepare 250ml of 0.500M NH3 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 14.8M NH3. Please prepare a flask containing 250ml of a 0.500M solution. To ensure proper credit, please write your calculations in the space given below.
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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.
See your other post. Note that C1V1 = C2V2 is the same as Ma x mLa = Mb x mLb
and note that 8.44 x 14.8 = 250 x 0.500 give or take a little for rounding
To prepare 250ml of a 0.500M NH3 solution, we will use the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the desired concentration, and V2 is the desired volume.
Given:
Initial concentration (C1) = 14.8M
Initial volume (V1) = 2.50L
Desired concentration (C2) = 0.500M
Desired volume (V2) = 250ml = 0.250L
We can rearrange the formula to solve for the initial volume:
V1 = (C2 * V2) / C1
Calculating the initial volume needed:
V1 = (0.500M * 0.250L) / 14.8M
V1 ≈ 0.00845L
Therefore, you will need to measure approximately 8.45ml of the 14.8M NH3 solution.
Here is a step-by-step procedure to prepare the 0.500M NH3 solution:
1. Measure approximately 8.45ml of the 14.8M NH3 solution using a graduated cylinder or pipette. Make sure to handle the concentrated solution cautiously and accurately transfer it to a 250ml volumetric flask.
2. Add distilled water gradually to the flask while swirling gently to ensure proper mixing. Continue adding water until the volume reaches the 250ml mark on the volumetric flask.
3. Once the solution is prepared, seal the flask with a stopper and invert it carefully several times to ensure homogeneity.
4. Label the flask with the concentration, date, and any other necessary information.
You have now successfully prepared a 250ml 0.500M NH3 solution using the 14.8M NH3 stock solution.
To prepare the 0.500M NH3 solution, we can use the equation C1V1 = C2V2, where C is concentration and V is volume.
First, let's determine the volume of the 14.8M NH3 stock solution needed. Since we have the concentrations and volumes for both the stock solution (C1 and V1) and the desired solution (C2 and V2), we can rearrange the equation as:
V2 = (C1 * V1) / C2
Given:
C1 = 14.8M
V1 = 2.50L
C2 = 0.500M
V2 = 0.250L
Now we substitute the given values into the equation:
V2 = (14.8M * 2.50L) / 0.500M
V2 = (37 Mol) / (0.500 M)
V2 = 74 L
So we need 74 liters of the 14.8M NH3 stock solution.
To prepare the 250ml of the 0.500M NH3 solution, we need to dilute the 14.8M solution to achieve the desired concentration. We can do this by dilution:
C1V1 = C2V2
Given:
C1 = 14.8M
V1 = 74L
C2 = 0.500M
V2 = 0.250L
Now we substitute the given values into the equation:
C1V1 = C2V2
(14.8M) * (74L) = (0.500M) * (0.250L)
(14.8M)(74L) = (0.500M)(0.250L)
1095.2 moles = 0.125 moles
Next, we can calculate the volume of the stock solution that needs to be diluted. We know that the number of moles of the substance before the dilution is the same as the number of moles after the dilution. Therefore:
moles before dilution = moles after dilution
moles before dilution = (concentration before dilution) * (volume before dilution)
moles after dilution = (concentration after dilution) * (volume after dilution)
Given:
moles before dilution = 1095.2 moles
concentration before dilution = 14.8M
Let's rearrange the equation to solve for the volume before dilution:
volume before dilution = moles before dilution / concentration before dilution
volume before dilution = 1095.2 moles / 14.8M
volume before dilution = 73.93 L
So we need to measure 73.93 liters of the 14.8M NH3 stock solution.
Finally, to prepare the 250ml of the 0.500M NH3 solution, we'll measure 250ml from the stock solution using a volumetric flask or a graduated cylinder. Make sure to accurately measure the 250ml mark on the flask. Then, add distilled water to the flask to reach the total volume of 250ml.
That's it! You have now prepared the 250ml of 0.500M NH3 solution for tomorrow's undergraduate experiment.