What volume of 6M HCl and 2M HCl should be mixed to get 2 litres of 3M HCl?
You either do it logically or by maths. Here is the maths solution.
If you need 2 L of 3M then you need a total of 6 moles.
Let x= volume of 6M and y=volume of 2M
so total moles is 6x+2y=6
total volume x+y=2
solve for x and y
I get x=0.5L and y=1.5L
To determine the volumes of 6M HCl and 2M HCl needed to obtain 2 litres of 3M HCl, we can use the formula:
C1V1 + C2V2 = C3V3
Where:
C1 = concentration of 6M HCl
V1 = volume of 6M HCl
C2 = concentration of 2M HCl
V2 = volume of 2M HCl
C3 = desired concentration (3M HCl)
V3 = desired volume (2 litres)
Substituting the given values into the equation, we have:
6M × V1 + 2M × V2 = 3M × 2L
Now, we can solve the equation step-by-step:
6V1 + 2V2 = 6
Since we are looking for a solution that has a total volume of 2 liters, we can assume that V1 + V2 = 2L. Rearranging the equation, we have V1 = 2L - V2.
Substituting the value of V1 into the equation, we get:
6(2L - V2) + 2V2 = 6
Expanding the equation:
12L - 6V2 + 2V2 = 6
Combining like terms:
12L - 4V2 = 6
Rearranging the equation to isolate V2:
4V2 = 12L -6
Dividing both sides by 4:
V2 = (12L - 6)/4
V2 = 3L - 1.5
We now have the value of V2, which represents the volume of 2M HCl needed. To find V1, we substitute this value into the equation V1 = 2L - V2:
V1 = 2L - (3L - 1.5)
Simplifying, we get:
V1 = 2L - 3L + 1.5
V1 = -L + 1.5
Therefore, to obtain 2 liters of 3M HCl, you would need to mix approximately -L + 1.5 liters (where L is any positive value, such as 1 liter) of 6M HCl with approximately 3 liters - (-L + 1.5) = L + 1.5 liters of 2M HCl.
To calculate the volumes of 6M HCl and 2M HCl required to obtain 2 liters of a 3M HCl solution, we can use the formula:
C1V1 + C2V2 = C3V3
Where:
C1 = concentration of the first solution
V1 = volume of the first solution
C2 = concentration of the second solution
V2 = volume of the second solution
C3 = desired concentration of the final solution
V3 = desired volume of the final solution
Let's plug in the values we know:
C1 = 6M
C2 = 2M
C3 = 3M
V3 = 2L
Now, we need to solve for V1 and V2. Rearranging the formula, we get:
V1 = (C3V3 - C2V2) / C1
V2 = (C3V3 - C1V1) / C2
Substituting the given values, we have:
V1 = (3M * 2L - 2M * V2) / 6M
V2 = (3M * 2L - 6M * V1) / 2M
Since we have two unknowns, we can use trial and error to find the values of V1 and V2 that satisfy both equations simultaneously. Here's one possible approach:
Assume V1 = 1L (for example)
V2 = (3M * 2L - 6M * 1L) / 2M
= (6M - 6M) / 2M
= 0L
In this case, the volume of the 2M HCl solution would be zero, meaning we only need to use the 6M HCl solution.
So, to make 2 liters of a 3M HCl solution, you would need 2 liters of 6M HCl and 0 liters of 2M HCl.