Question

A `5*g` mass of sodium hydroxide is dissolved in a volume of water such that the final solution is `1*L`...what are the `pOH`, and `pH` of this solution, and what is the final concentration of `H_3O^+`?

Answer 1

`pH=13.1`, `[H_3O^+]=10^(-13.1)*mol*L^-1=`

`[H_3O^+]=7.94xx10^-14*mol*L^-1`

Explanation:

We interrogate the equilibrium:

`2H_2O(l) rightleftharpoons H_3O^(+) + HO^-`

(Note that we may treat `H_3O^+` and `H^+` equivalently; this really is a matter of personal preference.)

We know, that this autoprotolysis reaction follows the given equilibrium reaction under standard conditions:

`2H_2O(l) rightleftharpoons H_3O^(+) + HO^-`

`K_w=[H_3O^+][""^(-)OH]=10^(-14)`

And if we take `log_10` of BOTH sides:

`log_10[H_3O^+]+log_10[HO^-]=-14`

EQUIVALENTLY, if we multiply each side by `-1`,

`14=-log_10[H_3O^+]-log_10[HO^-]`

But by definition, `-log_10[H_3O^+]=pH`, and..........

`-log_10[HO^-]=pOH`, and so............

`14=pH+pOH`, under standard conditions, and this is our defining relationship.

And so, (FINALLY after all that background):

`[HO^-]=(5*g)/(40*g*mol^-1)=0.125*mol`.

And `-log_10(0.125)=0.90=pOH`.

And thus `pH=14-pOH=13.1`

`[H_3O^+]=7.94xx10^-14*mol*L^-1`