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The [H3O+] for 0.26 M HCl is 0.26 Calculate the [OH-] for 0.26 M HCl. Answer in units of M.

Sagot :

We can assume that the concentration of H+ is 0.26M, because we are given [H3O+] and it is a strong acid.

So we will calculate the pH to obtain the pOH. Then apply pOH =-log[OH-]

pH = -log[H3O+]

pH = -log[0.26]

pH = 0.585

Then, we know that pH + pOH = 14

re-arrange:

pOH = 14 - pH

pOH = 14-0.585

pOH = 13.415

Now we can calculate the [OH-]

pOH = -log [OH-]

[OH-] = 10^-pOH

[OH] = 10^-13.415

[OH-] = 3.846x10^-14 M