25.0g of 5.0% (by mass) acetic acid solution are titrated with 0.300 M NaOH. What volume of NaOH will be needed to neutralize this sample?
25/molmassaceticacid * .05=.3*volumeinliters
To find the volume of NaOH needed to neutralize the acetic acid solution, we need to use the equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we know that the molar 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, let's convert the mass of acetic acid in the solution to moles:
Mass of acetic acid = 25.0 g
Percent by mass of acetic acid = 5.0%
To calculate the mass of acetic acid in the solution, we use the formula:
Mass of acetic acid = Mass of solution × Percent by mass of acetic acid / 100
Mass of acetic acid = 25.0 g × 5.0% / 100 = 1.25 g
To convert grams of acetic acid to moles, we use the molar mass of acetic acid, which is 60.05 g/mol:
Moles of acetic acid = Mass of acetic acid / Molar mass of acetic acid
Moles of acetic acid = 1.25 g / 60.05 g/mol ≈ 0.0208 mol
Since the molar ratio between acetic acid and sodium hydroxide is 1:1, we know that 0.0208 moles of acetic acid react with 0.0208 moles of sodium hydroxide.
Now, we can use the molar concentration (Molarity) of the NaOH solution and the balanced equation to find the volume of NaOH required.
Molarity of NaOH solution = 0.300 M
Using the equation:
Moles = Molarity × Volume
We can rearrange the equation to solve for the volume:
Volume of NaOH = Moles / Molarity
Volume of NaOH = 0.0208 mol / 0.300 mol/L ≈ 0.0693 L
Finally, we can convert the volume from liters to milliliters:
Volume of NaOH = 0.0693 L × 1000 mL/L ≈ 69.3 mL
Therefore, approximately 69.3 milliliters of NaOH will be needed to neutralize this sample.