What is the #pH# of a solution for which #[H_3O^+]=6.31xx10^-4*mol*L^-1#? Organic Chemistry Acids and Bases pH, pKa, Ka, pKb, Kb 1 Answer anor277 Jul 30, 2016 #pH=3.20# Explanation: #pH=-log_10[H_3O^+]# #=-log_10(6.31xx10^-4)=# #-log_10{6.31xx10^-4}=-(-3.20)=3.20# Answer link Related questions What is pH in acids and bases? What are pKa and pKb in acids and bases? What are Ka and Kb in acids and bases? How do you calculate the pH of a buffer solution? How can I calculate the pH of a weak acid with an example? How does Ka change with concentration? How does Ka relate to acid strength? How does molarity affect Ka? How does pH relate to #H^+#? How does pH relate to pKa? See all questions in pH, pKa, Ka, pKb, Kb Impact of this question 2146 views around the world You can reuse this answer Creative Commons License