A titration of\(50.0\;mL\)of unknown\(F{e^{2 + }}\)with\(0.100MCe\)at\(2{5^\circ }C\), monitored with Pt and calomel electrodes, gave data in the table.\(^9\)Prepare a Gran plot and decide which data lie on a straight line. Find the x-intercept of this line, which is the equivalence volume. Calculate the molarity of\(F{e^{2 + }}\)in the unknown.

• Iron (as Fe²⁺, ferrous ion) is a trace metal that is required by all known living species.
• Iron-containing enzymes, which usually have heme prosthetic groups, are involved in the catalysis of oxidation processes and the transport of a variety of soluble gases in biology.

A gran plot must be created for the given data, and the data must be forecasted to lie on a straight line.

This line's X-intercept must be discovered.

The level of concentration \(F{e^{2 + }}\) It must be calculated.

Given Data

To draw: gran plot for given data

The gran plot of \(V.1{0^{ - E0.059}}Vs\) V illustrated in Figure.1

The data between 8.50 and 12.50mL appears to be on a straight line.

To predict: this line's x-intercept

The slope of the least squares lines connecting these four points is m.

\(m = - 1.5673 \times 1{0^{ - 11}}\)

An intercept \(b = 2.1702 \times 1{0^{ - 10}}\)

x is the intercept.,

\(x = - \frac{b}{m} = \frac{{2.1702 \times 1{0^{ - 10}}}}{{\left( { - 1.5673 \times 1{0^{ - 11}}} \right)}} = 13.85\;mL\)

Calculating the concentration of \(F{e^{2 + }}\)

The amount of \(C{e^{4 + }}\) required to reach the point of equivalence is,

\((0.100mmol/mL)(13.85\;mL) = 1.385mmol\)

The amount of unidentified \(F{e^{2 + }}\) is,

\(1.385mmol/50.0\;mL = 0.0277M\)

A gran plot was created for the given data, and data lying on a straight line was forecasted as data ranging from 8.50 to 12.50mL.

This line's x-intercept was measured to be 13.85 mL.

\(F{e^{2 + }}\)concentration was calculated to be 0.0277 M.