## Chemistry 9th Edition

a. $[H^+]=[OH^-]=1.71*10^{-7}\text{ M}$ b. $pH =6.77$ c. $pH=12.5$
a. $[H^+]=[OH^-]=x$ in pure water. Therefore, $[H^+][OH^-]=2.92*10^{-14}$ $x^2=2.92*10^{-14}$ $x=1.71*10^{-7}\text{ M}$ $[H^+]=[OH^-]=1.71*10^{-7}\text{ M}$ b. $pH = -log(1.71*10^{-7})=6.77$ c. $[H^+][OH^-]=2.92*10^{-13}$ $0.10*[H^+]=2.92*10^{-13}$ $[H^+]=2.92*10^{-12}$ $pH=-log[2.92*10^{-12}]=12.5$