How do you find the pH, the pOH, [H3O+], and [OH-] in equations?

Question

Wyatt Atkins · Accepted Answer

#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#

#K_a = [H_3O^+][HO^-] =10^(-14)# #K_a = [H_3O^+][HO^-] =10^(-14)# This ion product, this equilibrium, has been established by experiment at #298# #K#. At higher temperatures, given that this is a bond-breaking reaction, how would you expect this equilibrium to evolve? Now, given the equation, we can manipulate it, as long as we do it to both sides of the equation. So we take #log_10# of BOTH sides to give: #log_10K_a# #=log_10[H_3O^+] +log_10[HO^-]#. And on rearrangement: #-log_10[H_3O^+] -log_10[HO^-]=-log_10K_a# But by definitions, #-log_10[H_3O^+]=pH#, #-log_10[HO^-]=pOH#, and #-log_(10)10^-14=14#. And thus #pH + pOH =14.# As there should be plenty of examples of problems on these boards, get to work. If you are at A level, all you need to know is this final result. However, you do need to be able to apply it to solve problems.

Luna Chen · Answer

undefined To find the pH, pOH, [H3O+], and [OH-] in equations, you can use the following formulas:

pH = -log[H3O+]
pOH = -log[OH-]
[H3O+] = 10^(-pH)
[OH-] = 10^(-pOH)

You can calculate these values using the concentrations of hydronium ions ([H3O+]) and hydroxide ions ([OH-]) in the solution.