   Chapter 3, Problem 16PS

Chapter
Section
Textbook Problem

Two aqueous solutions were prepared, one containing 0.10 mol of boric acid (H3BO3) in 200 mL. and the second containing 0.10 mol phosphoric add (H3PO4) in 200 mL. Both were weak conductors of electricity, but the H3PO4 solution was a noticeably stronger conductor. Write equations to describe the equilibrium in each solution, and explain the observed difference in conductivity.

Interpretation Introduction

Interpretation:

The equation that explains an equilibrium in each solution has to be written and observed conductivity difference should be explained.

Explanation

The equation for the phosphoric acid at equilibrium is given below, phosphoric acid is tribasic acid it can donate the proton to the water molecule and conduct the electricity.

H3PO4+ H2H2PO4-+ H3O+H2PO4-+ H2OHPO42-+ H3O+HPO42-+ H2OPO43-+ H3O+

The equation for the boric acid in aqueous solution is given below, boric acid is can donate the proton to the water molecule and conduct the electricity

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 