What is the molarity of an HCl solution if 49.0 mL is completely titrated by 68.4 mL of an NaOh solution whose concentration is 0.333 M?
To find the molarity of the HCl solution, we can use the equation:
M1V1 = M2V2
Where:
M1 = molarity of NaOH solution
V1 = volume of NaOH solution used
M2 = molarity of HCl solution
V2 = volume of HCl solution used
Given:
M1 = 0.333 M
V1 = 68.4 mL
V2 = 49.0 mL
First, we need to convert the volumes to liters by dividing by 1000:
V1 = 68.4 mL / 1000 mL/L = 0.0684 L
V2 = 49.0 mL / 1000 mL/L = 0.0490 L
Now we can plug in the values into the equation:
M1V1 = M2V2
(0.333 M)(0.0684 L) = M2(0.0490 L)
Simplifying:
0.0223 = 0.0490 M2
Now, solving for M2:
M2 = 0.0223 / 0.0490
M2 ≈ 0.4549 M
Therefore, the molarity of the HCl solution is approximately 0.4549 M.
To find the molarity (M) of the HCl solution, we can use the equation:
M₁V₁ = M₂V₂
Where:
M₁ = Molarity of the HCl solution
V₁ = Volume of the HCl solution in liters
M₂ = Molarity of the NaOH solution
V₂ = Volume of the NaOH solution in liters
First, let's convert the volumes from milliliters (mL) to liters (L):
V₁ = 49.0 mL ÷ 1000 = 0.049 L
V₂ = 68.4 mL ÷ 1000 = 0.0684 L
Now we can plug in the values into the equation:
M₁(0.049 L) = (0.333 M)(0.0684 L)
Rearranging the equation to solve for M₁:
M₁ = (0.333 M)(0.0684 L) ÷ 0.049 L
Now we can calculate the molarity of the HCl solution:
M₁ = (0.333)(0.0684) ÷ 0.049
M₁ ≈ 0.462 M
Therefore, the molarity of the HCl solution is approximately 0.462 M.