Question

Using the systematic approach for equilibrium problems, calculate the pH of 0.05 M HOCl. Ka= 3.0*10-8 Group of answer choices 3.87 7.62 1.00 4.41

Answer 1

The pH is equal to 4.41

Step-by-step explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

HClO ⇄ ClO- + H+

start: 0.05 0 0

change -x +x +x

balance 0.05-x x x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

`Ka=([ClO-]*[H+])/([HClO])=(x*x)/(0.05-x)=3x10^(-8)`

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

`3x10^(-8)=(x^(2) )/(0.05)`

clearing the x and calculating its value we have:

`x=3.87x10^(-5)=[H+]=[ClO-]`

the pH can be calculated by:

`pH=-log[H+]=-log[3.87x10^(-5)]=4.41`