Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677e: 98

$N=0.082\ mol$

Initial pH: $pH=pKa+\log([A^-]/[HA])$ $[A^-]=[HA]\rightarrow pH=pKa=4.76$ Case of adding strong acid: $pH'=pH-1=3.76$ $3.76=4.76+\log([A^-]/[HA])$ $0.1=[A^-]/[HA]$ $[A^-]=(0.10\ M\times 1L-N)/V_f$ $[HA]=(0.10\ M\times 1L+N)/V_f$ $0.1=(0.1-N)/(0.1+N)$ $0.01+0.1N=0.1-N$ $0.09=1.1N$ $N=0.082\ mol$