A #0.45*g# mass of #HCl# is dissolved in water to give a #300*mL# volume of solution. What are #pOH#, and #[HO^-]# for this solution?

1 Answer
Mar 22, 2017

Answer:

We interrogate the equilibrium:

#HCl(g) + H_2O(l)rightleftharpoons H_3O^+ +Cl^-#

Explanation:

#HCl(g) + H_2O(l)rightleftharpoons H_3O^+ +Cl^-#

This equilibrium lies almost entirely to the right, and we assume the solution is stoichiometric in #H_3O^+#.

And thus #[HCl]=((0.45*g)/(36.46*g*mol^-1))/(300.0xx10^-3L)=0.041*mol*L^-1#.

Because dissociation is QUANTITATIVE, #[H_3O^+]=[HCl]=0.041*mol*L^-1#.

And #pH=-log_10[H_3O^+]=1.39#

#pOH=14-pH=12.61#, and thus #[HO^-]=10^(-12.61)="something small"#