Assuming the density of a 5% acetic acid solution is 1g/ml, determine the volume of the acetic acid solution necessary to neutralize 25ml of 0.1m naoh also record this calculation on your report
My calculation
1g/ml x 1000ml x 0.05/61 = 0.81M
0.81m = 25ml x 0.1
0.81m = 2.5
x= 2.5/0.81
x= 3ml
or 25ml x .01m = 2.5ml x 61 /0.05/ 1
x 3ml
Who helps me to solve it for me
To solve the problem, you can use the formula:
(Concentration of acetic acid) x (Volume of acetic acid) = (Concentration of NaOH) x (Volume of NaOH)
Let's calculate the volume of the acetic acid solution needed step-by-step:
Given:
Concentration of acetic acid = 5% = 0.05
Density of acetic acid solution = 1 g/mL
Volume of NaOH = 25 mL
Concentration of NaOH = 0.1 M
Step 1: Calculate the molarity of the acetic acid solution.
Multiply the density of the acetic acid solution by the percent concentration and divide by the molar mass of acetic acid (61 g/mol):
Molarity = (1 g/mL) x (1000 mL) x (0.05/61) = 0.81 M
Step 2: Set up the equation using the molarity values.
(0.81 M) x (Volume of acetic acid) = (0.1 M) x (25 mL)
Step 3: Solve for the volume of acetic acid.
Volume of acetic acid = (0.1 M x 25 mL) / 0.81 M
Volume of acetic acid = 2.5 mL / 0.81
Volume of acetic acid = 3.09 mL (rounded to two decimal places)
So, you would need approximately 3.09 mL of the 5% acetic acid solution to neutralize 25 mL of 0.1 M NaOH.
You can include these steps in your report to show your calculation process.
To solve the problem, you can use the equation:
M₁V₁ = M₂V₂
Where
M₁ is the molarity of the acetic acid solution,
V₁ is the volume of the acetic acid solution needed,
M₂ is the molarity of the sodium hydroxide (NaOH) solution, and
V₂ is the volume of the sodium hydroxide solution given.
In this case,
M₂ = 0.1 M (given)
V₂ = 25 mL (given)
First, we need to calculate the molarity (M₁) of the acetic acid solution using the density provided.
Density = mass/volume
Since density is given as 1 g/mL, that means 1 mL of the solution weighs 1 gram.
Convert the percentage concentration to molarity by using the molecular weight of acetic acid (CH₃COOH), which is 61 g/mol.
Molarity (M₁) = (1 g/mL × 1000 mL × 0.05) / 61
Molarity (M₁) ≈ 0.081 M (rounded to three decimal places)
Now, substitute the values into the equation:
0.081 M × V₁ = 0.1 M × 25 mL
Solve for V₁:
V₁ ≈ (0.1 M × 25 mL) / 0.081 M
V₁ ≈ 30.86 mL (rounded to two decimal places)
Therefore, the volume of the acetic acid solution necessary to neutralize 25 mL of 0.1 M NaOH is approximately 30.86 mL.