How much water must be added to 500.mL of .200M HCL to produce a .150M solution?
(.500mL x .200M)/.150M = .667mL
Correct?
A chemistry student needs 125mL f .150M NaOH solution for her experiment, but the only solution avail. is 3.00M. Describe how the student could prepare the solution needed.
I don't know about this one
Your first one is close to being right. I just asked my chem teacher and he said that the V is the volume of the solution, not just the solvent. Therefore, you should subtract the already existing 500ml of water from those 667ml of solution, giving you 167ml of water needed.
For the first one, check your work. I think you have an error of about 1000 times. I think 500*0.200/0.150 = 667 mL. However, that is the volume you will have, not how much water must be added. Water that must be added is 667-500 = ??
For the second one,
M*mL = M*mL.
125 mL * 0.15 M = 3.0 M x mL NaOH.
mL 3.0 M NaOH = (125*0.15/3.0) = 6.25 mL of the 3.0 M NaOH.
So measure exactly 6.25 mL of the 3.0 M NaOH and make to exactly 125 mL. (All of that may be easier said than done. Measuring exactly 6.25 mL could be done rather accurately with a buret (not many pipets of 6.25 mL) BUT making to exactly 125 mL would be hard to do for there aren't any 125 mL volumetric flasks. At least I've not seen one. But there could be a way around that. The student could add 12.50 mL of the 3.0 M NaOH to a 250 mL volumetric flask, then add water to the mark. That would be exact.
Well, if the student wanted to end up with 125mL of a 0.150M NaOH solution but only has a 3.00M solution available, they could use some good ol' dilution!
They could take 10 mL of the 3.00M NaOH solution and add 115 mL of water to it. This would give them a total volume of 125 mL (10 mL + 115 mL) and a final concentration of approximately 0.150M (0.3 M x [(10 mL/125 mL)] = 0.150M).
Remember, it's like the solution is saying, "Hey, I'm too strong, I need to be watered down a bit!" And the student says, "Sure, let's dilute you with some water!" And then they both live happily ever after. La la la!
To calculate the amount of water needed to dilute a solution, you can use the dilution formula:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
For the first question, you want to dilute 500 mL of a 0.200 M HCl solution to a final concentration of 0.150 M. We can use the dilution formula to find the volume of water needed.
C1 = 0.200 M
V1 = 500 mL
C2 = 0.150 M
V2 = unknown (we'll solve for this)
Plugging in the values into the formula:
(0.200 M)(500 mL) = (0.150 M)(V2)
Solving for V2:
V2 = (0.200 M)(500 mL) / (0.150 M)
V2 = 666.67 mL
Therefore, approximately 666.67 mL of water must be added to the 500 mL of 0.200 M HCl to produce a 0.150 M solution.
For the second question, the student has a 3.00 M NaOH solution but needs a 0.150 M NaOH solution. To prepare the desired solution, the student can perform a dilution by adding water.
Since the student needs 125 mL of a 0.150 M solution, they can use the dilution formula again:
C1 = 3.00 M
V1 = unknown (we'll solve for this)
C2 = 0.150 M
V2 = 125 mL
Plugging in the values into the formula:
(3.00 M)(V1) = (0.150 M)(125 mL)
Solving for V1:
V1 = (0.150 M)(125 mL) / (3.00 M)
V1 = 6.25 mL
Therefore, the student can measure 6.25 mL of the 3.00 M NaOH solution and dilute it with water to a final volume of 125 mL to prepare the desired 0.150 M NaOH solution.
To find out how much water must be added to the HCl solution to obtain a 0.150M concentration, you can use the dilution formula:
C₁V₁ = C₂V₂
Where:
C₁ is the initial concentration (0.200M)
V₁ is the initial volume (500 mL)
C₂ is the final concentration (0.150M)
V₂ is the final volume (unknown)
Rearranging the formula to solve for V₂, we have:
V₂ = (C₁ × V₁) / C₂
Substituting the given values:
V₂ = (0.200M × 500 mL) / 0.150M
V₂ = 1.333 mL
So, the correct answer is 1.333 mL of water should be added to the 500 mL of 0.200M HCl solution to obtain a 0.150M solution.
Regarding the second question, if a chemistry student needs 125 mL of a 0.150M NaOH solution but only has a 3.00M solution available, they can calculate the amount of the concentrated solution required and then dilute it with water.
Using the dilution formula mentioned earlier:
C₁V₁ = C₂V₂
Where:
C₁ is the initial concentration (3.00M)
V₁ is the initial volume (unknown)
C₂ is the final concentration (0.150M)
V₂ is the final volume (125 mL)
Rearranging the formula to solve for V₁, we have:
V₁ = (C₂ × V₂) / C₁
Substituting the given values:
V₁ = (0.150M × 125 mL) / 3.00M
V₁ = 6.25 mL
So, the student should measure 6.25 mL of the 3.00M NaOH solution and then add enough water to reach a total volume of 125 mL to obtain the desired 0.150M NaOH solution.