A mixture of 0.0500 moles of each of the following is prepared in 1.000 L of DI

Question

A mixture of 0.0500 moles of each of the following is prepared in 1.000 L of DI water:

HCl (large Ka)

CH3COOH, acetic acid (Ka = 1.8x10-5)

C6H5OH, phenol (Ka = 1.6x10-10)

1. Account for all activity coefficients and for the changes caused by the solution having an ionic strength that should not be neglected by including the appropriate activity coefficients, determine the concentrations of the species listed below at equilibrium in the solution:     HCl, Cl-, H3O+, CH3COOH, CH3COO-, C6H5OH, C6H5O-, and OH- (also find pH).

Explanation / Answer

>>In presence of a strong acid like HCl, acetic acid and phenol will be in their protonated, i.e. undissociated form.
Hence concentration of CH3COO- and C6H5O- will be zero. And that of CH3COOH and C6H5OH will be 0.0500 moles/ Lt.
HCl will be completely dissociated and so its concept. Will be zero. And concept. of Cl- and H3O+ will be 0.0500 moles/ Lt.

As [H3O+][OH-] = 10^-14
Therefore [OH-] = 10^-14/0.0500 = 20 X 10^-14

pH = - log[H+] = - log(0.0500) = 1.30

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.