Answer
a. $[H^+]=[OH^-]=1.71*10^{-7}\text{ M}$
b. $pH =6.77$
c. $pH=12.5$
Work Step by Step
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.
$[H^+][OH^-]=2.92*10^{-13}$
$0.10*[H^+]=2.92*10^{-13}$
$[H^+]=2.92*10^{-12}$
$pH=-log[2.92*10^{-12}]=12.5$
