## Chemistry 9th Edition

a. $[OH^-]=1.0*10^{-7}\text{ M}$, Neutral b. $[OH^-]=1.2*10^1\text{ M}$, Basic c. $[OH^-]=8.2*10^{-16}\text{ M}$, Acidic d. $[OH^-]=1.9*10^{-10}\text{ M}$, Acidic
a. $[H^+][OH^-]=1.0*10^{-14}$ $1.0*10^{-7}[OH^-]=1.0*10^{-14}$ $[OH^-]=1.0*10^{-7}\text{ M}$ $pH = -log[H^+]=-log(1.0*10^-7)=7$ Therefore, the solution is neutral. b. $[H^+][OH^-]=1.0*10^{-14}$ $8.3*10^{-16}[OH^-]=1.0*10^{-14}$ $[OH^-]=1.2*10^1\text{ M}$ Because $[OH^-]>[H^+]$, the solution is basic. c. $[H^+][OH^-]=1.0*10^{-14}$ $12[OH^-]=1.0*10^{-14}$ $[OH^-]=8.2*10^{-16}\text{ M}$ Because \$[OH^-]