Question

A 0.0712-g sample of a purified organic acid was dissolved in an alcohol-water mixture and titrated with coulometrically generated hydroxide ions. With a current of 0.0392 A, 241 s was required to reach a phenolphthalein end point. Calculate the equivalent mass of the acid.

Short answer

The equivalent mass of the acid is 727.27 g/equiv.

Step-by-step solution

Given Information

Mass of sample = 0.0712 g

Current = 0.0392 A

Time = 241 s

Explanation

The expression of the quantity of charge by using Faraday law is:

$Q=It$…… (I)

Here, the constant current is I and the time period is t.

The expression of the equivalent mass of acid is:

${\mathrm{m}}_{\mathrm{a}}=\frac{{\mathrm{m}}_{\mathrm{HA}}}{\mathrm{Q}}$…… (III)

Here, the mass of sample of purified organic acid is m_HA.

Substitute 0.0392 A for I and 241 s for t in Equation (I).

$$\begin{gathered} \mathrm{Q}=(0.0392\mathrm{A})\left(241\mathrm{s}\right) \\ =(0.0392\mathrm{A})\left(\frac{1\mathrm{C}/\mathrm{s}}{1\mathrm{A}}\right)\left(241\mathrm{s}\right) \\ =(9.45\mathrm{C})\left(\frac{1\mathrm{F}}{96485\mathrm{C}}\right)\left(\frac{1\mathrm{equiv}\mathrm{HA}}{1\mathrm{F}}\right) \\ =(9.79\times {10}^{-5}\mathrm{equiv}\mathrm{HA}) \end{gathered}$$

Substitute 9.79 x 10^-5 equiv HA for Q and 0.0712 g HA for m_HA in Equation (II).

$$\begin{gathered} {\mathrm{m}}_{\mathrm{a}}=\frac{\left(0.0712\mathrm{g}\mathrm{HA}\right)}{\left(9.79\times {10}^{-5}\mathrm{equiv}\mathrm{HA}\right)} \\ =\frac{\left(0.0712\mathrm{g}\right)}{\left(9.79\times {10}^{-5}\mathrm{equiv}\right)} \\ =727.27\mathrm{g}/\mathrm{equiv} \end{gathered}$$

Conclusion

727.27 g/equiv is the equivalent mass of the acid.