Question

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?

Answer 1

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"`