How much 12 M HCl do you add to 1 L of water to obtain 0.2 HCl? Use this same approach to do the math for the dilution of 5.5 M of barium hydroxide solution to 0.5 M barium hydroxide solution.
Do you mean 0.2 M HCl? I assume so.
Use the dilution formula for both HCl and for Ba(OH)2.
That is c1v1 = c2v2
c = concn
v = volume
To determine the amount of 12 M HCl needed to obtain a 0.2 M HCl solution, we can use the dilution formula:
C₁V₁ = C₂V₂
Where:
- C₁ is the initial concentration of the HCl solution (12 M)
- V₁ is the initial volume of the HCl solution (unknown)
- C₂ is the final concentration of the HCl solution (0.2 M)
- V₂ is the final volume of the HCl solution (1 L)
Rearranging the formula, we have:
V₁ = (C₂V₂) / C₁
Substituting the values, we get:
V₁ = (0.2 M * 1 L) / 12 M
V₁ = 0.0167 L
Therefore, you would need to add approximately 0.0167 liters (or 16.7 milliliters) of 12 M HCl to 1 liter of water to obtain a 0.2 M HCl solution.
Now let's use the same approach to calculate the dilution of 5.5 M barium hydroxide (Ba(OH)₂) solution to 0.5 M.
Using the same dilution formula:
C₁V₁ = C₂V₂
Where:
- C₁ is the initial concentration of the barium hydroxide solution (5.5 M)
- V₁ is the initial volume of the barium hydroxide solution (unknown)
- C₂ is the final concentration of the barium hydroxide solution (0.5 M)
- V₂ is the final volume of the barium hydroxide solution (unknown)
Rearranging the formula, we have:
V₁ = (C₂V₂) / C₁
Let's say we want a final volume of 1 L (V₂ = 1 L), we can substitute this value:
V₁ = (0.5 M * 1 L) / 5.5 M
V₁ = 0.0909 L
Therefore, you would need to take approximately 0.0909 liters (or 90.9 milliliters) of the 5.5 M barium hydroxide solution and dilute it with water to obtain a 0.5 M barium hydroxide solution with a final volume of 1 liter.