Answer: 10⁻⁸°⁹
Step-by-step explanation:
[tex]pH=log\bigg(\dfrac{1}{[H^+]}\bigg)\\\\\text{use log expansion rules to get:}\\pH=log\ 1-log\ [H^+]\\\\8.9=0\ -\ log\ [H^+]\\\\8.9=-log\ [H^+]\\\\-8.9=log\ [H^+]\\\\\text{use log transformation rules to get the following exponential equation:}\\10^{-8.9}=[H^+][/tex]
