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^+#?

1 Answer
Apr 4, 2017

Answer:

#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#