a. A student obtains a vinegar solution having an acetic acid concentration of 0

Question

a. A student obtains a vinegar solution having an acetic acid concentration of 0.8512 M. What volume of 0.1036 M NaOH will be required to reach the equivalence point when 25.00 mL of the vinegar solution is titrated?

b. A student dilutes the vinegar solution in part a above by pipetting 10 mL of the solution into a 100 mL volumetric flask and diluting to a total volume of 100 mL. What volume of 0.1036 M NaOH will be required to reach the equivalence point when 100 mL of the diluted vinegar solution is titrated?

c. What volume of 0.1036 M NaOH will be required to reach the equivalence point when 25 mL of the diluted vinegar solution prepared in part b is titrated?  

Explanation / Answer

acetic acid = CH3COOH

CH3COOH + NaOH ? CH3COONa + H2O

(0.8512 mol/L CH3COOH) x (0.025 L) x (1 mol NaOH/1 mol CH3COOH) /
(0.103 mol/L NaOH) = 0.207 L NaOH

Yes there is a formula. (And it's difficult to believe you've never seen it if you're being asked to work this problem.)
It is:
v1 x c1 = v2 x c2
where the v's are volumes and the c's are concentrations. Expressed this way, this formula only works for acids and bases that react in equimolar amounts.
The problem at hand uses this formula in this arrangement:
v2 = v1x c1 / c2

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