Question

A titration of 50.0 mLof unknown ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$with 0.100MCeat $\mathbf{25}\mathbf{°}\mathbf{C}$ , monitored with Pt and calomel electrodes, gave data in the table. 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 ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$in the unknown.

Short answer

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. Concentration was calculated to be 0.0277 M.

Step-by-step solution

Definition of Fe2+

• 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.

Determine the x-intercept of this line, which is the equivalence volume and molarity of Fe2+ in the unknown.

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 It must be calculated.

Given Data

To draw: gran plot for given data

The gran plot of $\mathrm{V}.{10}^{-\mathrm{E}0.059}\mathrm{Vs}\mathrm{V}$ 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.

$\mathrm{m}=-1.5673\times {10}^{-11}$

An intercept

x is the intercept.,

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

Calculating the concentration of

The amount of required to reach the point of equivalence is,

$\left(0.100\mathrm{m}\mathrm{mol}/\mathrm{mL}\right)\left(13.85\mathrm{m}\mathrm{L}\right)=1.385\mathrm{m}\mathrm{mol}$

The amount of unidentified is,

1.385mmol/50.0mL = 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.

concentration was calculated to be 0.0277 M.