posted by Amy .
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 the 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.
Technically, this problem can't be worked unless you give an equation or you know the ratio of acid to base. I ASSUME it is either NaOH + HCl ==> NaCl+ H2O OR
NaOH + HC2H3O2 ==> NaC2H3O2 + H2O in which case the mole ratio is 1:1 in either case.
mols acid used = M x L = 0.10 x 0.008 L = ??
Since the mol ratio is 1:1, then mols base used is the same.
Molarity base = mols/L.
You know mols and you know liters. Solve for molarity.
For the second part, use the concept shown in the problem above to logically answer the question but I will give you a hint. The indicator you are uses changes color when mols acid = mols base. Will the volume change affect the number of moles of either?
thank you. for the second part the increase in volume should not affect the number of moles right? so would i put no change?
That is exactly right. The volume has no effect on the end point. I remember very well when I did demonstrations to show students how solutions were titrated. Of course I used a wash bottle to periodically wash the sides down and I would wait for a student to ask, and they always did every year, doesn't the addition of wash water dilute the material in the beaker? And I always answered, yes it does but it also dilutes the material from the buret, too, by exactly the same amount. Then I would go through the mols routine. When students realize that its the mols vs mols, not concentration versus concentration, they understand.