Question

The ionization constant of HF is 3.2 X 10—4. Calculate the degree of dissociation of HF in its 0.02M solution. Calculate the concentration of all the species present in the solution and its PH.? please it's urgent.

Answer 1

Answer: Degree of dissocaition = 0.016

Concentration of $H^+$= 0.00032 M

Concentration of $F^-$= 0.00032 M

Concentration of $HF$= 0.019 M

pH= 3.5

Explanation:

Given: ionization constant of HF =$3.2\times 10^{-4}$

Concentration of HF= 0.02 M

To calculate the moles, we use the formula:

          $HF\longrightarrow H^++F^-$

initially      c         0     0

at eq'm $c- c\alpha$ $c\alpha$   $c\alpha$

So ionization constant will be:

$K_{a}=\frac{c\alpha\times c\alpha}{c-c\alpha}$

$3.2\times 10^{-4}=\frac{0.02\alpha\times 0.02\alpha}{0.02-0.02\alpha}$

$\alpha=0.016$

Thus degree of dissociation= 0.016

Concentration of $H^+$ at equilibrium =$c\alpha=0.02\times 0.016=0.00032M$

$pH=-log[H^+]=-log[0.0032]=3.5$

Concentration of $F^-$ at equilibrium =$c\alpha=0.02\times 0.016=0.00032M$

Concentration of $HF$ at equilibrium =$c-c\alpha=0.02-0.02\times 0.016=0.019M$

Answer 2

Explanation:

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