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^+#?

Question

Caleb Adams · Accepted Answer

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

#[H_3O^+]=7.94xx10^-14*mol*L^-1# We scrutinize the balance: #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 are aware that, in typical circumstances, this autoprotolysis reaction yields the specified equilibrium reaction: #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. Thus, (after all that background, FINALLY): #[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#

Wyatt Atkins · Answer

undefined To find the pOH of the solution, use the equation: pOH = -log[OH-]. Since NaOH is a strong base, it dissociates completely in water, so [OH-] = concentration of NaOH. 
pOH = -log(5 mol/L) = -log(5) ≈ 0.7.

Since pH + pOH = 14, pH = 14 - pOH.
pH = 14 - 0.7 = 13.3.

The concentration of H3O+ in a basic solution can be calculated using the Kw expression: Kw = [H3O+][OH-]. 
Kw = 1.0 x 10^-14 (at 25°C), and [OH-] = 5 mol/L (from the given NaOH concentration).
[H3O+] = Kw / [OH-] = (1.0 x 10^-14) / 5 = 2.0 x 10^-15 M.