What volume of 0.149 M HCl must be added to 1.00x10^2 mL of 0.285 M HCl so that the resulting solution has a molarity of 0.205 M? Assume that the volumes are additive.
Let x = volume in mL of 0.149 M HCl to be added and mmols = millimols.
mmols 0.285M + mmols 0.149M = mmols 0.205M
(0.285*100) + (0.149x) = 0.205*(100+x)
Solve for x.
You should prove that by adding mmols 0.285M to mmols 0.149M and see if total mmols/total volume in mL = 0.205M
Well, if we're adding HCl, it appears that we're in a rather acidic situation. Good thing I'm here to bring a little laughter to balance things out!
To solve this problem, we can use the concept of Molarity and the equation: M1V1 = M2V2. M1 and V1 represent the initial molarity and volume, while M2 and V2 represent the final molarity and volume, respectively.
Given that M1 = 0.285 M, V1 = 1.00x10^2 mL, M2 = 0.205 M, and we need to find V2, we can set up the equation as:
(0.285 M)(1.00x10^2 mL) = (0.205 M)(V2)
Solving for V2, we get:
V2 = (0.285 M)(1.00x10^2 mL) / 0.205 M
Now, let's do some calculations: (Make some silly noises while doing math)
V2 = 139.02 mL
So, to achieve a molarity of 0.205 M, you would need to add 139.02 mL of 0.149 M HCl. It seems like a tiny amount compared to the funny reactions I've seen in chemistry labs! Enjoy your acid balance!
To solve this problem, we can use the concept of molarity and volume relationships. The formula for calculating molarity is:
Molarity (M) = moles of solute / volume of solution (in liters)
In this case, we are given three pieces of information:
1. Initial volume of 0.285 M HCl solution: 1.00 x 10^2 mL
2. Final volume of resulting solution: This is not specified, but we can assume it to be the sum of the initial volume and the volume of 0.149 M HCl to be added.
3. Target molarity of resulting solution: 0.205 M
Let's solve the problem step by step:
Step 1: Convert the initial and final volumes to liters:
Initial volume of 0.285 M HCl solution: 1.00 x 10^2 mL = 1.00 x 10^2 / 1000 = 0.100 L
Final volume of resulting solution: We'll denote this as V (in liters), which is not specified.
Step 2: Determine the moles of HCl in the initial solution:
Molarity (M) = moles / volume
0.285 M = moles / 0.100 L
moles = 0.285 M x 0.100 L = 0.0285 moles
Step 3: Determine the moles of HCl needed in the final solution:
We can use the equation:
Molarity (M) = moles / volume
0.205 M = (0.0285 moles + x moles) / (0.100 L + V)
Step 4: Solve for the unknown moles and volume:
Rearranging the equation from step 3:
0.205 M * (0.100 L + V) = 0.0285 moles + x moles
0.0205 L + 0.205V = 0.0285 + x
0.205V = 0.0285 - 0.0205 + x
0.205V = 0.008 + x
Step 5: Convert the moles of 0.149 M HCl to volume in liters:
We can use the equation:
moles = Molarity (M) x volume (V)
Volume = moles / Molarity
Volume = 0.008 / 0.149 M ≈ 0.05369 L
Therefore, to obtain a resulting solution with a molarity of 0.205 M, you need to add approximately 0.05369 L or 53.69 mL of 0.149 M HCl to the initial 1.00 x 10^2 mL of 0.285 M HCl solution.
To find the volume of 0.149 M HCl that must be added, we can set up an equation based on the principle of dilution, which states that the moles of solute before and after dilution should be the same.
Let's break down the problem step by step:
Step 1: Calculate the number of moles of HCl in the initial solution.
We have 1.00 x 10^2 mL of 0.285 M HCl. To find the number of moles, we use the formula:
moles = concentration (M) × volume (L)
Converting the volume from mL to L:
1.00 x 10^2 mL = 1.00 x 10^2 / 1000 = 0.100 L
Now calculate the number of moles of HCl:
moles = 0.285 M × 0.100 L = 0.0285 moles
Step 2: Now, let's use the equation for dilution:
moles_initial = moles_final
moles_initial refers to the moles of HCl initially in the solution, and moles_final refers to the moles of HCl after adding the additional HCl.
For the final solution to have a molarity of 0.205 M, we need to find the total number of moles:
moles_final = 0.205 M × (0.100 L + x)
Here, x represents the volume of 0.149 M HCl that needs to be added in liters.
Now set up the equation:
0.0285 moles = 0.205 M × (0.100 L + x)
Step 3: Solve the equation for x.
0.0285 moles = 0.205 M × 0.100 L + 0.205 M × x
0.0285 = 0.0205 + 0.205x
0.008 = 0.205x
x = 0.008 / 0.205
x ≈ 0.039 L
Step 4: Convert the volume back to milliliters.
x = 0.039 L × 1000 = 39 mL
Therefore, approximately 39 mL of the 0.149 M HCl solution needs to be added to achieve a resulting solution with a molarity of 0.205 M.