If the endpoint, in the titration of the KHC8H404 with NaOH, is accidentally surpassed (too pink), what effect will this have on the calculated molarity of the NaOH solution?

1 Answer
Nov 18, 2015

Answer:

See explanation.

Explanation:

The reaction that is taking place is the following:

#HC_8H_4O_4^(-)(aq)+OH^(-)(aq)->C_8H_4O_4^(2-)(aq)+H_2O(l)#

From the reaction, we can tell that the number of mole of #HC_8H_4O_4^(-)# is equal to the number of mole of #OH^-#:

#n_(HC_8H_4O_4^(-))=n_(OH^-)#

#=>(C_MxxV)_(HC_8H_4O_4^(-))=(C_MxxV)_(OH^-)#

#=>(C_M)_(OH^-) =((C_MxxV)_(HC_8H_4O_4^(-)))/V_(OH^-)#

Therefore, the molarity of the #NaOH# is inversely proportional to the volume of #NaOH# added.

Thus, when #V_(NaOH)# increases , #(C_M)_(NaOH) # will decrease.

Here is a video that features an experimental demonstration similar to this topic:

Lab Experiment #9: Determination of the Formula Mass of an Acid Salt.