There were 0.04 mol of HA in 100 mL of solution.
You didn't have to do a titration, since you knew the concentration of the original solution.
You started with 110 mL of 0.5 mol/L HA and diluted it to a total of 150mL.
To calculate the concentration of the dilute solution, you can use the dilution formula
The moles of HA in 100 mL of this solution are
Note: The answer can have only 1 significant figure, because that is all you gave for the concentration of the HA. If you need more precision, you will have to recalculate.
The reaction for the titration was
HA + NaOH → NaA + H₂O
The moles of HA in the sample you titrated were
But this was for 10.00 mL of solution. In 100 mL of solution there would be
Note: The answer can have only 1 significant figure, because that is all you gave for the concentration of the NaOH. If you need more precision, you will have to recalculate.
The answer is indeed
So, you know that you had 110.0 mL of 0.5 M acetic acid. You figure out the number of moles of acetic acid present in this sample
The acetic acid solution, the one with the added 40.0 mL of water, will have 0.055 moles of acetic acid, because adding water will not affect the number of moles of acetic acid already present, it will only affect the solution's concentration.
You can use a simple proportion to figure out how many moles of acetic acid you have in a 100-mL sample of this 150.0-mL solution