Question

I could use some help filling in the RICE table. Especially the Reaction row. I can solve the problem after that part.

Practice Exercise 15.19 What is the difference in pH between a 0.25 M solution of dimethyl ammonium chloride, (CH3)2NH2Cl, and a 0.25 M NH4CI solution? REACTION Initial Change Equilibriunm

Answer 1

For the solution 0.25 M NH₄Cl:

The NH₄Cl dissociates like this:

$NH_{4}Cl\Leftrightarrow NH_{4}^{+}+Cl^{-}$

The one that hydrolyzes is the ammonium ion because it comes from a weak base (ammonium), this is:

$NH_{4}^{+}+H_{2}O\Leftrightarrow NH_{3}+H_{3}O^{+}$

For the solution 0.25 M (CH₃)₂NH₂Cl:

The (CH₃)₂NH₂Cl dissociates like this:

$\left ( CH_{3} \right )_{2} NH\Left_{2}Cl \Leftrightarrow \left ( CH_{3} \right )_{2}NH_{2}^{+}+Cl^{-}$

The one that hydrolyzes is the diethylammonium ion because it comes from a weak base, this is:

$\left ( CH_{3} \right )_{2}NH_{2}^{+}+H_{2}O\Leftrightarrow \left ( CH_{3} \right )_{2}NH+H_{3}O^{+}$