## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677a: 26

#### Answer

It goes from 9.21 to 9.34

#### Work Step by Step

Initially, all hydroxide added is neutralized: Number of moles: $20/1000\ L\times 0.10\ M=0.002\ mol$ $OH^-+NH_4^+\rightarrow NH_3 +H_2O$ The new concentration of ammonium and ammonia are: $NH_4^+: (0.183\ M\times 80.0/1000\ L-0.002\ mol)\div 100/1000\ L=0.126\ M$ $NH_3: (0.169\ M\times 80.0/1000\ L+0.002\ mol)\div 100/1000\ L=0.155\ M$ New equilibrium: $K_a=x(0.155+x)/(0.126-x)$ $x=4.55\cdot10^{-10}\ M=[H_3O^+]$ $pH=9.34$ From Henderson-Hasselbach, the original pH was 9.21, So it goes from 9.21 to 9.34

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.