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

Answer:

The pH is equal to 4.41

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:

$Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}$

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

$3x10^{-8}=\frac{x^{2} }{0.05}$

clearing the x and calculating its value we have:

$x=3.87x10^{-5}=[H+]=[ClO-]$

the pH can be calculated by:

$pH=-log[H+]=-log[3.87x10^{-5}]=4.41$