assuming the density of a 5% acetic acid solution is 1.0 g/mL, determine the volume of the acetic acid solution necessary to neutralize 25.0mL of 0.10 M Na)H. Also record this calculation on your Report Sheet.
5% acetic acid solution means 5 g acetic acid in 100 g soln.
If the density of the soln is 1 g/mL, then the soln has a volume of 100 mL and the acetic solution is 50 g/1000 mL.
Convert 50 g CH3COOH to moles and you will have moles/L = M.
To determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH, we can use the equation:
Molarity (M) × Volume (V) = Molarity (M) × Volume (V)
We know the following values:
Molarity of acetic acid solution (M1) = 5%
Density of acetic acid solution = 1.0 g/mL
Volume of NaOH solution (V2) = 25.0 mL
Molarity of NaOH solution (M2) = 0.10 M
First, we need to convert the percentage into molarity:
Molarity (M) = Percentage/100
Molarity (M1) = 5/100 = 0.05 M
Next, we need to convert the volume of the NaOH solution into liters:
Volume (V2) = 25.0 mL = 25.0 mL × (1 L/1000 mL) = 0.025 L
Now we can substitute the values into the equation:
(Molarity of acetic acid solution) × (Volume of acetic acid solution) = (Molarity of NaOH solution) × (Volume of NaOH solution)
(0.05 M) × (Volume of acetic acid solution) = (0.10 M) × (0.025 L)
Now we can solve for the volume of acetic acid solution:
Volume of acetic acid solution = (0.10 M × 0.025 L) / 0.05 M
Volume of acetic acid solution = 0.0025 L / 0.05
Volume of acetic acid solution = 0.05 L
The volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH is 0.05 L or 50 mL.
Make sure to record this calculation on your Report Sheet.
To determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH, we need to use the concept of stoichiometry and the balanced equation for the neutralization reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH).
The balanced equation for this reaction is:
CH3COOH + NaOH → CH3COONa + H2O
According to the balanced equation, the stoichiometric ratio between CH3COOH and NaOH is 1:1. This means that for every 1 mole of CH3COOH, we need 1 mole of NaOH.
Given that the concentration of NaOH solution is 0.10 M and the volume is 25.0 mL, we can determine the number of moles of NaOH using the formula:
moles = concentration x volume
moles of NaOH = 0.10 mol/L x 0.025 L = 0.0025 moles
Since the stoichiometric ratio between CH3COOH and NaOH is 1:1, we know that we would need 0.0025 moles of CH3COOH to neutralize the given amount of NaOH.
Now, let's calculate the mass of CH3COOH required using its molar mass. The molar mass of CH3COOH is approximately 60.05 g/mol.
mass = moles x molar mass
mass of CH3COOH = 0.0025 moles x 60.05 g/mol = 0.150 g
Since we have the density of the acetic acid solution as 1.0 g/mL, we can now calculate the volume of the acetic acid solution required:
volume = mass / density
volume of acetic acid solution = 0.150 g / 1.0 g/mL = 0.150 mL
Therefore, the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH is approximately 0.150 mL.
Make sure to record this calculation on your report sheet.