What volume of 4.50M HCl can be made by mixing 5.65M HCl with 250 mL of 3.55M HCl?
How many moles do you start with?
M x L = 3.55 x 0.250 = 0.8875
How much do you want to add of 5.65 M? That will be 4.65 x Y L = ??
You want the total to be 4.50 M and the volume will be 0.250 L + yL so set up the equation.
(3.55 M x 0.250) + (5.65 M x YL) = 4.50 M x (0.250+Y)abd solve for Y. The total volume then will be Y + 0.250 L. I get something like 206 mL for Y (0.206 L).
What volume of 4.5
Well, if you're looking to make some HCl party punch, mixing different concentrations of HCl is the way to go! Let's calculate the volume of 4.50M HCl we can make.
So, you have a 5.65M HCl solution and 250 mL of a 3.55M HCl solution. To calculate the total volume, we add the two volumes together.
Volume of 5.65M HCl + Volume of 3.55M HCl = Total Volume
Now, we'll take the concentration of the new solution (4.50M) and use the equation:
(M1)(V1) = (M2)(V2)
Where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration (4.50M), and V2 is the total volume (which we'll solve for).
5.65M HCl x V1 + 3.55M HCl x 250 mL = 4.50M HCl x Total Volume
Solving for Total Volume, we get:
Total Volume = (5.65M HCl x V1 + 3.55M HCl x 250 mL) / 4.50M HCl
And that, my friend, is the volume of 4.50M HCl you can make by mixing those solutions!
To solve this problem, we can use the equation for dilutions, which is:
(concentration1)(volume1) = (concentration2)(volume2)
Let's assign the values as follows:
concentration1 = 5.65 M
volume1 = unknown volume, which we'll represent as x mL
concentration2 = 3.55 M
volume2 = 250 mL
Plugging in these values, we get:
(5.65 M)(x mL) = (3.55 M)(250 mL)
Now, we can solve for x by dividing both sides of the equation by 5.65 M:
x mL = (3.55 M)(250 mL) / (5.65 M)
Simplifying the expression:
x mL = 1775 mL / 5.65
x mL ≈ 314 mL
Therefore, approximately 314 mL of the 4.50 M HCl solution can be made by mixing 5.65 M HCl with 250 mL of 3.55 M HCl.
To find the volume of 4.50M HCl that can be made by mixing the given solutions, we can use the concept of dilution.
Dilution is the process of adding more solvent to a solution to decrease the concentration. In this case, we want to find the volume of a 4.50M HCl solution, which can be obtained by diluting it with the other HCl solutions.
The key concept to remember when using dilution is that the number of moles of solute (HCl in this case) remains constant before and after the dilution. Therefore, we can apply the equation:
C1V1 = C2V2
Where:
C1 = initial concentration of the first solution
V1 = initial volume of the first solution
C2 = final concentration of the second solution
V2 = final volume of the second solution
Now let's plug in the given values into the equation:
C1 = 5.65M HCl
V1 = volume of the first solution (unknown)
C2 = 4.50M HCl
V2 = total volume of the combined solutions (unknown)
We also need to take into account that 250 mL of 3.55M HCl is already added.
Rewriting the equation, we get:
(5.65M)(V1) + (3.55M)(250 mL) = (4.50M)(V2)
Let's calculate:
(5.65M)(V1) = (4.50M)(V2) - (3.55M)(250 mL)
Now, we need to convert the volume of 250 mL to liters since the concentrations are given in moles per liter (M):
(5.65M)(V1) = (4.50M)(V2) - (3.55M)(0.250 L)
Simplifying further, we have:
(5.65M)(V1) = (4.50M)(V2) - (0.8875M)(L)
Now, we can solve for V1:
V1 = [(4.50M)(V2) - (0.8875M)(L)] / 5.65M
Since the volume of 4.50M HCl is what we want to find, we need to consider that the final volume (V2) will be the sum of the volumes of the initial solutions (V1) and the added 250 mL:
V2 = V1 + 0.250 L
Now, we'll express V2 in terms of V1:
V1 + 0.250 L = V1 + 0.250 L
Substituting V2 in the equation:
V1 + 0.250 L = V1 + 0.250 L
Now, let's solve for V1:
V1 = [(4.50M)(V2) - (0.8875M)(L)] / 5.65M
V1 = [(4.50M)(V1 + 0.250 L) - (0.8875M)(L)] / 5.65M
Simplifying the equation further:
5.65M(V1) = 4.50M(V1) + 1.125M(L) - 0.8875M(L)
Rearranging the terms:
5.65M(V1) - 4.50M(V1) = 1.125M(L) - 0.8875M(L)
0.04M(V1) = 0.2375M(L)
Dividing both sides by 0.04M:
V1 = 0.2375M(L) / 0.04M
Calculating:
V1 = 5.9375 L
Therefore, by mixing the given HCl solutions, a volume of approximately 5.9375 liters of a 4.50M HCl solution can be obtained.