Answer: Concentration of [tex][H^+]\&[OH^-][/tex] are equal in pure water.
Explanation:
[tex]H_2O\rightleftharpoons H^++OH^-[/tex]
[tex]K_{eq}=\frac{[H^+][OH^-]}{[H_2O]}[/tex]
[tex]K_w=K_{eq}[H_2O]=[H^+][OH^-]=1.0\times 10^{-14}[/tex]
Also, pure water has neutral with 7 pH value ,
[tex]pH=7=-\log[H^+][/tex]
[tex][H^+]=1\times 10^{-7}mol/L[/tex]
So, the[tex]K_w [/tex] of pure water is given
[tex]K_w=[H^+][OH^-]=1\times 10^{-7}\times[OH^-]= 1\times 10^{-14}[/tex]
[tex][OH^-]=1\times 10^{-7}mol/L[/tex]
Hence, Concentration of [tex][H^+]\&[OH^-] [/tex]are equal in pure water
