How much 0.10 M Base (in mL) is required to neutralize 8.00 mL of the 0.10 M acid?

Base Acid Base Volume (mL)
NaOH HCl ___________
NaOH HC2H3O2 ___________

How would the volume of base change from problem above if the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?

A. The volume would decrease.
B. No change.
C. The volume would increase.
D. Not enough information is given to predict what would happen.

i don't know why but the rest of my question isnt coming up

t=1f=q

This question is answered above.

the question is not answered above!! please help!!

To find the volume of 0.10 M base required to neutralize 8.00 mL of the 0.10 M acid, we can use the concept of stoichiometry and the balanced chemical equation.

Let's assume the balanced chemical equation for the neutralization reaction is:
Acid + Base -> Salt + Water

Now, we need to use the stoichiometry of the balanced equation to find the volume of base required. According to the balanced equation, the molar ratio between the acid and the base is 1:1. This means that for every mole of acid, we need 1 mole of base to neutralize it.

First, let's find the number of moles of acid in 8.00 mL of the 0.10 M acid:
moles of acid = volume of acid (in L) x concentration of acid (in mol/L)
= 8.00 mL x (1 L / 1000 mL) x 0.10 mol/L
= 0.008 mol

Since the molar ratio between acid and base is 1:1, we need 0.008 moles of base to neutralize the acid.

Now, let's find the volume of 0.10 M base required to provide 0.008 moles:
volume of base = moles of base / concentration of base
= 0.008 mol / 0.10 mol/L
= 0.08 L
= 80 mL

So, 80 mL of the 0.10 M base is required to neutralize 8.00 mL of the 0.10 M acid.

Now, let's consider the second part of the question regarding the dilution of the acid. When the 8.00 mL of 0.10 M acid is diluted with 100 mL of water, the total volume of the acid solution becomes 8.00 mL + 100 mL = 108.00 mL.

However, it's important to note that the concentration of the acid remains the same, 0.10 M. This is because dilution refers to adding more solvent (water in this case) to the solution without changing the amount of solute (the acid).

Therefore, since the concentration of the acid is unchanged, the volume of the base required to neutralize it would still be 80 mL.

The answer to the question is B. No change.