What is $p H$ $pH$ if $[H_{3} O^{+}] = 0.001 \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=0.001molL^-1$ ?

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Answer 1 · 2017-06-08T11:48:38

$pH=3............$

$pH$ , $"pouvoir hydrogene"$ , $=-log_10[H_3O^+]$

Now it is a fact that in water the following autoprotolysis takes place:

$2H_2O(l) rightleftharpoonsHO^(-) + H_3O^+$

The ionic product at $298*K$ $=$ $K_w=[H_3O^+][HO^-]=10^-14$ .

If we take $-log_10$ of both sides we get:

$pK_w =14=-log_10[H_3O^+]-log_10[HO^-]$

But by definition, $pH =-log_10[H_3O^+]$ and $pOH =-log_10[HO^-]$

Thus $pK_w =14=pH+pOH$

But nitric acid should be stoichiometric in $H_3O^+$ and $NO_3^-$ ion...........

$HNO_3(aq) + H_2O(l) rarr H_3O^+ + NO_3^-$

And so $pH$ $=$ $-log_10(10^-3)$ $=$ $3$ , and $pOH=??$

And $pH=3$ .

Capisce?

What is pHpHpH if [H_3O^+]=0.001*mol*L^-1[H3O+]=0.001⋅mol⋅L−1[H_3O^+]=0.001*mol*L^-1?