# 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?

Nov 18, 2015

See explanation.

#### Explanation:

The reaction that is taking place is the following:

$H {C}_{8} {H}_{4} {O}_{4}^{-} \left(a q\right) + O {H}^{-} \left(a q\right) \to {C}_{8} {H}_{4} {O}_{4}^{2 -} \left(a q\right) + {H}_{2} O \left(l\right)$

From the reaction, we can tell that the number of mole of $H {C}_{8} {H}_{4} {O}_{4}^{-}$ is equal to the number of mole of $O {H}^{-}$:

${n}_{H {C}_{8} {H}_{4} {O}_{4}^{-}} = {n}_{O {H}^{-}}$

$\implies {\left({C}_{M} \times V\right)}_{H {C}_{8} {H}_{4} {O}_{4}^{-}} = {\left({C}_{M} \times V\right)}_{O {H}^{-}}$

$\implies {\left({C}_{M}\right)}_{O {H}^{-}} = \frac{{\left({C}_{M} \times V\right)}_{H {C}_{8} {H}_{4} {O}_{4}^{-}}}{V} _ \left(O {H}^{-}\right)$

Therefore, the molarity of the $N a O H$ is inversely proportional to the volume of $N a O H$ added.

Thus, when ${V}_{N a O H}$ increases , ${\left({C}_{M}\right)}_{N a O H}$ 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.