What is the pOH of a 0.001 M HCl solution?

Question

What is the pOH of a 0.001 M HCl solution?

1 Answer

Impact of this question

7912 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-15T07:02:35

Well, we know that $p H + p O H = 14$ $pH+pOH=14$ , so..............

$p O H = 11$ $pOH=11$

Explanation:

For the autoprotolysis reaction:

$2 H_{2} O (l) ⇌ H_{3} O^{+} + H O^{-}$ $2H_2O(l) rightleftharpoons H_3O^+ +HO^(-)$ ,

we know that $K_{w} = [H_{3} O^{+}] [H O^{-}] = 10^{- 14}$ $K_w=[H_3O^+][HO^-]=10^-14$ under standard conditions............

And taking ${log}_{10}$ $log_10$ of both sides we gets............

${log}_{10} K_{w} = {log}_{10} 10^{- 14} = {log}_{10} [H_{3} O^{+}] + {log}_{10} [H O^{-}]$ $log_10K_w=log_(10)10^-14=log_10[H_3O^+]+log_10[HO^-]$

And thus $- {log}_{10} [H_{3} O^{+}] - {log}_{10} [H O^{-}] = - {log}_{10} 10^{- 14} = 14$ $-log_10[H_3O^+]-log_10[HO^-]=-log_(10)10^(-14)=14$

But by definition, $- {log}_{10} [H_{3} O^{+}] = p H$ $-log_10[H_3O^+]=pH$ , and $- {log}_{10} [H O^{-}] = p O H$ $-log_10[HO^-]=pOH$

So................

$p H + p O H = 14$ $pH+pOH=14$

You will not be asked to reproduce this as a first year student or as an A-level student. You will be asked (as you have been here!) to use the relationship.

And $[H O^{-}] = 10^{- 11} \cdot m o l \cdot L^{- 1}$ $[HO^-]=10^(-11)*mol*L^-1$ .

Science

Math

Humanities

... and beyond

What is the pOH of a 0.001 M HCl solution?

1 Answer

Explanation:

Related questions

Impact of this question