How do you calculate the pOH of a solution?

Question

How do you calculate the pOH of a solution?

1 Answer

Impact of this question

1901 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-06T05:38:45

In water you use the relationship $p H + p O H = 14$ $pH+pOH=14$ .

Explanation:

We know that at $298 \cdot K$ $298*K$ , equilibrium obtains between hydronium ion and hydroxide ions in water:

$K_{w} = 10^{- 14} = [H O^{-}] [H_{3} O^{+}]$ $K_w=10^(-14)=[HO^-][H_3O^+]$ . We can take $- {log}_{10}$ $-log_10$ of both sides:

$- {log}_{10} (K_{w}) = - {log}_{10} [H O^{-}] - {log}_{10} [H_{3} O^{+}] = - {log}_{10} (10^{- 14})$ $-log_10(K_w)=-log_10[HO^-]-log_10[H_3O^+]=-log_10(10^-14)$

i.e. $- {log}_{10} [H O^{-}] - {log}_{10} [H_{3} O^{+}] = 14$ $-log_10[HO^-]-log_10[H_3O^+]=14$ by definition of the log function. And again by definition $p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^+]$ etc.

Thus $p H + p O H = 14$ $pH+pOH=14$ . There should be many problems here that use this relationship. Of course, you have to find them. Good luck.

Science

Math

Humanities

... and beyond

How do you calculate the pOH of a solution?

1 Answer

Explanation:

Related questions

Impact of this question