If the endpoint in the titration of KHC8H4O4 solution with the NaOH solution is mistakenly surpassed, will the molar concentration of NaOH solution be reported too high or too low?
If the endpoint in the titration of KHC8H4O4 solution with the NaOH solution is mistakenly surpassed, the molar concentration of NaOH solution will be reported too high.
If the endpoint in the titration of KHC8H4O4 solution with the NaOH solution is surpassed, it means that more NaOH has been added than necessary to neutralize the KHC8H4O4 solution. As a result, the molar concentration of NaOH solution will be reported too high.
To understand why this happens, let's discuss the concept behind titration. In a titration, a known concentration of a solution (called the titrant) is slowly added to a solution of unknown concentration (called the analyte) until a chemical reaction reaches completion. This completion point is known as the endpoint.
In this case, we are titrating a known concentration of NaOH solution to determine the concentration of KHC8H4O4 solution. The reaction between NaOH and KHC8H4O4 is acid-base neutralization, where the acid (KHC8H4O4) reacts with the base (NaOH) to form water and a salt (potassium salt of the acid).
KHC8H4O4 + NaOH -> H2O + K+ + C8H4O4-
The stoichiometry of this reaction is 1:1, meaning that one mole of KHC8H4O4 reacts with one mole of NaOH. Therefore, if the endpoint is surpassed, it means that excess NaOH has been added to the reaction mixture.
When the endpoint is surpassed, the excess NaOH reacts with the indicator or causes a sharp change in pH, indicating that the reaction is now complete. However, this excess NaOH is not required to neutralize the KHC8H4O4 solution and contributes to the total volume of the NaOH solution used.
Since the molar concentration of a solution is calculated by dividing the moles of solute by the volume of solution used, surpassing the endpoint by adding excess NaOH will result in a larger volume of NaOH solution used. This increase in volume will cause an overestimation of the concentration of the NaOH solution and, consequently, the molar concentration will be reported too high.
(1) mols KHC8H4O4 = grams/molar mass
(2)mols NaOH = mols KHC8H4O4 (since the mols ratio is 1:1 in the titration. Write and balance the equation to confirm that.)
(3) M NaOH = mols NaOH/L NaOH.
So you over titrate with the base. No mL are involved in equation 1 above. No change. No mL involved in equation 2. No change. But mL NaOH are involved in equation 3, that makes a larger denominator which make the answer smaller than it should be.