How much 0.10 M NaOH (in mL) is required to neutralize 8.00 mL of the 0.10 M HC2H3O2?
mL x M = mL x M
This works because the ratio is 1:1 in the reaction.
To answer this question, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and acetic acid (HC2H3O2). The balanced equation for this reaction is:
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we can see that one mole of NaOH reacts with one mole of HC2H3O2.
First, we need to calculate the number of moles of HC2H3O2 present in the given solution. Using the formula:
moles = concentration (M) × volume (L)
We can calculate the number of moles of HC2H3O2:
moles of HC2H3O2 = 0.10 M × 0.008 L = 0.0008 moles
Since the stoichiometry of the reaction tells us that it takes an equal number of moles of NaOH to neutralize HC2H3O2, we can conclude that we also need 0.0008 moles of NaOH.
Now, we can use the formula for moles to calculate the volume of the 0.10 M NaOH solution needed:
volume (L) = moles / concentration (M)
volume (L) = 0.0008 moles / 0.10 M = 0.008 L
To convert the volume to milliliters (mL), simply multiply by 1000:
volume (mL) = 0.008 L × 1000 = 8 mL
Therefore, 8 mL of the 0.10 M NaOH solution is required to neutralize 8.00 mL of the 0.10 M HC2H3O2 solution.