calculate the molarity of an acetic acid solution if 25.00mL of the solution required 35.25mL of 0.1200M NaOH for neutralization.
mL NaOH x M NaOH = mL HAc x M HAc
Well, let's dive into this a-bit-ic acid problem! We need to calculate the molarity, right?
First, we can look at the balanced equation:
CH3COOH + NaOH → CH3COONa + H2O
Now, the mole ratio is 1:1, my friend. This means 1 mole of acetic acid reacts with 1 mole of sodium hydroxide. So, the number of moles of acetic acid that reacted is the same as the number of moles of sodium hydroxide.
We have:
0.1200 M NaOH = 35.25 mL NaOH
Now, we can calculate the moles of NaOH:
moles of NaOH = volume (in liters) × concentration
moles of NaOH = 35.25 mL × 0.1200 mol/L = 0.004230 mol
Since the ratio is 1:1, the moles of acetic acid are also 0.004230 mol.
To find the molarity, we use:
molarity = moles of solute / volume (in liters) of solution
Here we go:
molarity = 0.004230 mol / (25.00 mL / 1000 mL/L)
molarity = 0.1692 M
So the Molarity of the acetic acid solution is 0.1692 M.
Remember, laughter is the best solution, whether diluted or concentrated!
To calculate the molarity (M) of an acetic acid solution, we can use the equation:
M₁V₁ = M₂V₂
Where:
M₁ = molarity of acetic acid solution (what we are trying to find)
V₁ = volume of acetic acid solution (25.00 mL)
M₂ = molarity of NaOH solution (0.1200 M)
V₂ = volume of NaOH solution (35.25 mL)
Now let's plug in the values into the equation and solve for M₁:
M₁ (25.00 mL) = (0.1200 M) (35.25 mL)
M₁ = (0.1200 M) (35.25 mL) / (25.00 mL)
M₁ = 0.04248 M
Therefore, the molarity of the acetic acid solution is 0.04248 M.
To calculate the molarity of the acetic acid solution, we can use the concept of stoichiometry and the equation of the balanced chemical reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH).
The balanced chemical equation for the reaction between acetic acid and sodium hydroxide is:
CH3COOH + NaOH -> H2O + CH3COONa
From the balanced equation, we can see that the ratio between acetic acid and sodium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
First, we need to determine the number of moles of NaOH used in the neutralization reaction. This can be calculated using the formula:
moles of NaOH = molarity of NaOH x volume of NaOH solution (in liters)
Given:
Molarity of NaOH = 0.1200 M (moles per liter)
Volume of NaOH solution = 35.25 mL = 35.25/1000 = 0.03525 L
moles of NaOH = 0.1200 M x 0.03525 L = 0.00423 moles
Since the ratio between acetic acid and NaOH is 1:1, the moles of acetic acid used in the reaction is also 0.00423 moles.
Now, we need to calculate the molarity of the acetic acid solution.
To do this, we use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
We have the moles of acetic acid, which is 0.00423 moles. However, we need to find the volume of the acetic acid solution in liters.
Given:
Volume of acetic acid solution = 25.00 mL = 25.00/1000 = 0.025 L
Molarity (M) = 0.00423 moles / 0.025 L = 0.1692 M
Therefore, the molarity of the acetic acid solution is 0.1692 M.