In a clean 250mL volumetric flask, 125mL of a 6.0 M hydrochloride acid was added. This was then diluted with more water to reach the 250mL mark. What is the molarity of the acid at this time?
The volume is doubled, concentration is halved.
To find the molarity of the acid after dilution, we can use the equation:
M₁V₁ = M₂V₂
Where:
M₁ = Initial molarity of the acid
V₁ = Initial volume of the acid
M₂ = Final molarity of the acid
V₂ = Final volume of the acid
In this case:
M₁ = 6.0 M (initial molarity of the acid)
V₁ = 125 mL (initial volume of the acid)
M₂ = ?
V₂ = 250 mL (final volume of the acid)
We can substitute these values into the equation and solve for M₂:
6.0 M × 125 mL = M₂ × 250 mL
To solve for M₂, we can rearrange the equation:
M₂ = (6.0 M × 125 mL) / 250 mL
Now let's calculate the molarity of the acid after dilution:
M₂ = (6.0 M × 125 mL) / 250 mL
= (6.0 × 125) / 2
= 750 / 2
= 375 / 1
= 375 M
Therefore, the molarity of the acid after dilution is 375 M.