A technician measures 1.75 L of a standard solution of hydrochloric acid, HCl(aq), with a concentration of 9.00 mol/L. Determine the amount of water that must be added to create a solution with a concentration of 2.50 mol/L.
mL1 x M1 = mL2 x M2
1.75 mL x 9.00M = mL2 x 2.50
Solve for mL2. That is the TOTAL volume needed. Technically you can't find the actual volume to be ADDED but it will be close if you subtract 1.75 L from the total. To make the solution accurately, however, you add the 1.75 L to a volumetric flask of the volume found for mL2 and add water to the final volume.
Well, let's think about it. If the concentration of the original hydrochloric acid solution is 9.00 mol/L and we want to end up with a concentration of 2.50 mol/L, we need to dilute it.
To dilute the solution, we need to add water. But wait a minute, if we add water, won't the solution get weaker? It's like adding water to a really strong cup of espresso - no one likes weak coffee!
Anyway, back to the problem. We can use the equation C1V1 = C2V2 to solve it. We know the initial concentration (C1) is 9.00 mol/L, the initial volume (V1) is 1.75 L, and the desired concentration (C2) is 2.50 mol/L. We need to find the final volume (V2) of the diluted solution.
Using the equation, we can rearrange it to get V2 = (C1V1) / C2. Plugging in the values, we get V2 = (9.00 mol/L * 1.75 L) / 2.50 mol/L.
Now, let's do some math. *cracks knuckles* Multiplying 9.00 mol/L by 1.75 L gives us 15.75 mol, and dividing that by 2.50 mol/L gives us a final volume of... 6.3 liters!
So, you need to add 6.3 liters of water to the 1.75 liters of hydrochloric acid to end up with a solution that has a concentration of 2.50 mol/L. Just remember, dilution is like weak coffee – no one likes it!
To determine the amount of water that must be added to create a solution with a concentration of 2.50 mol/L, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration (9.00 mol/L)
V1 = initial volume (1.75 L)
C2 = final concentration (2.50 mol/L)
V2 = final volume (unknown)
Rearrange the formula to solve for V2:
V2 = (C1V1) / C2
Substitute the known values:
V2 = (9.00 mol/L * 1.75 L) / 2.50 mol/L
V2 = 6.75 L
Therefore, the technician must add 6.75 liters of water to create a solution with a concentration of 2.50 mol/L.
To solve this problem, we need to use the dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case, the initial concentration (C1) is 9.00 mol/L and the initial volume (V1) is 1.75 L. The final concentration (C2) is 2.50 mol/L, and we need to find the final volume (V2).
First, let's rearrange the formula to solve for V2:
V2 = (C1 * V1) / C2
Now, substitute the given values:
V2 = (9.00 mol/L * 1.75 L) / 2.50 mol/L
Simplify the expression:
V2 = 15.75 L / 2.50 mol/L
V2 = 6.3 L
The final volume (V2) is 6.3 liters.
To determine the amount of water that must be added, we subtract the initial volume (V1) from the final volume (V2):
Amount of water = V2 - V1
Amount of water = 6.3 L - 1.75 L
Amount of water = 4.55 L
Hence, 4.55 liters of water must be added to create a solution with a concentration of 2.50 mol/L.