A concentration of at least 5 p.p.m - Leaving Cert Chemistry - Question 1 - 2019

A large excess of KI was required to ensure complete reduction of iodine produced during the titration. The presence of excess iodide ions helps to shift the equilibrium of the reaction in favor of the formation of free iodide ions, which allows for accurate measurement of the iodine produced from the oxidation of Mn$^{2+}$ ions.

The conical flask was rinsed with deionised water to remove any contaminants. After rinsing, a measured volume of the sample solution was added to the flask, ensuring it was accurately filled to the desired level. During the titration, a white tile was placed beneath the flask to improve visibility of the color change. The titration was performed carefully, adding the titrant until a distinct color change was observed, indicating the endpoint.

Initially, the color of the solution changed from brown to pale yellow as iodine was consumed. The solution became progressively lighter until it nearly cleared, signaling the reduction of iodine. Upon adding starch as an indicator at the final stages, the solution turned blue-black, indicating the presence of free iodine prior to reaching the endpoint.

To calculate the amount of sodium thiosulfate consumed during the titration, the concentration of Na$_2$S$_2$O$_3$.5H$_2$O is needed. First, we must find the moles of sodium thiosulfate used based on its molar mass and the volume used in titrations. Using the formula:

$M = \frac{m}{M_f}$

where m is mass and $M_f$ is the molar mass, we can calculate the moles for the solution used.