How many liters of water must be added to 4 liters of a 6 molar HNO3 solution to create a solution that is 2 molar?
You think 4 L = 0.004 mL? Think again, once to get the correct volume, then a second time after you have re-read my note.
Sorry i got confused so it is 4000 mL. So will the final volume be the L that is added?
No, the final volume will be the final volume. You STARTED with 4000 mL and the final volume is xxx; therefore, xxx - 4000 will be the mL that must be added.
Technically, that isn't exactly right because the volumes usually are not additive but I think the intent of this problem is to assume the volumes ARE additive.
To solve this problem, we need to use the concept of dilution.
Dilution formula:
M1V1 = M2V2
Where:
M1 = initial concentration (molarity) of the solution (HNO3 in this case)
V1 = initial volume of the solution (4 liters of HNO3 in this case)
M2 = final concentration (molarity) of the solution (2 M in this case)
V2 = final volume of the solution (we need to find this)
Let's substitute the given values into the equation and solve for V2:
(6 M) * (4 L) = (2 M) * (4 L + V2)
24 = 8 + 2V2
Subtracting 8 from both sides of the equation:
2V2 = 24 - 8
2V2 = 16
Dividing both sides of the equation by 2:
V2 = 8
Therefore, you need to add 8 liters of water to 4 liters of a 6 M HNO3 solution to create a solution that is 2 M.
So will it be 0.004*6=2(mL)
0.024/2=mL
0.012mL ?
mL x M = mL x M
(Note: the final volume will NOT be the number of mL that must be added.)