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

Question

1842 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-04T19:30:20

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

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

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 $pH=14-pOH=13.1$

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

A 5*g5*g mass of sodium hydroxide is dissolved in a volume of water such that the final solution is 1*L1*L...what are the pOHpOH, and pHpH of this solution, and what is the final concentration of H_3O^+H_3O^+?