What is pH if [H_3O^+]=0.001*mol*L^-1?

Jun 8, 2017

$p H = 3. \ldots \ldots \ldots . .$

Explanation:

$p H$, $\text{pouvoir hydrogene}$, $= - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

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

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s H {O}^{-} + {H}_{3} {O}^{+}$

The ionic product at $298 \cdot K$ $=$ ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$.

If we take $- {\log}_{10}$ of both sides we get:

$p {K}_{w} = 14 = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right]$

But by definition, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$

Thus $p {K}_{w} = 14 = p H + p O H$

But nitric acid should be stoichiometric in ${H}_{3} {O}^{+}$ and $N {O}_{3}^{-}$ ion...........

$H N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + N {O}_{3}^{-}$

And so $p H$ $=$ $- {\log}_{10} \left({10}^{-} 3\right)$ $=$ $3$, and pOH=??

And $p H = 3$.

Capisce?