Question

If `pH=2.6` in aqueous solution, what is `pOH`?

Answer 1

We know that `pH+pOH=14` under standard conditions....

Explanation:

Now `pOH=-log_10[HO^-]...`

And here `pOH=-log_10(0.0025)=-(-2.60)=2.60`

And so `pH=14-2.60=11.4`

As to background, just to note that in aqueous solution under standard conditions, the ion product...

`K_w=[H_3O^+][HO^-]=10^(-14)`...

And we can take `log_10` of both sides to give....

`log_(10)K_w=log_(10)10^(-14)=log_10[H_3O^+]+log_10[HO^-]`.

And thus.... `-14=log_(10)[H_3O^+]+log_(10)[HO^-]`

Or.....

`14=-log_(10)[H_3O^+]-log_(10)[HO^-]`

`14=underbrace(-log_10[H^+])_(pH)underbrace(-log_10[OH^-])_(pOH)`

`14=pH+pOH`

And so here by definition, `-log_10[H^+]=pH`, `-log_10[HO^-]=pOH`

Answer 2

pH =11.40

Explanation:

We know that `"NaOH"` is a strong electrolyte. It dissociates completely in solution.

`color(white)(mmmmmml)"NaOH(aq)" → "Na"^"+""(aq)" + "OH"^"-""(aq)"`
`"I/mol·L"^"-1": color(white)(mm)0.0025 color(white)(mmmmm)0color(white)(mmmmll)0`
`"C/mol·L"^"-1": color(white)(m)"-0.0025"color(white)(mmm)"+0.0025"color(white)(mll)"+0.0025"`
`"E/mol·L"^"-1": color(white)(mmm)0color(white)(mmmmm)0.0025color(white)(mml)0.0025`

Thus, the concentration of `"OH"^"-"` is 0.0025 mol/L.

`"pOH = -log"["OH"^"-"] = "-log(0.0025) = 2.60"`

`"pH = 14.00 - pOH = 14.00 - 2.60 = 11.40"`