Question

Que. 5. Dissociation constant of NH4OH is 1.8  10-5. Calculate its degree of

dissociation in 0.01M solution.​

Answer 1

Dissociation constant, $K_{b} = 1.8$ × 10⁻⁵

          Concentration , C = 0.01M

The degree of dissociation of a weak electrolyte is directly proportional to the extent of dilution(square root of dilution)

The formula to calculate the degree of dissociation is

                 ∝ = $\sqrt{\frac{K_{b} }{C} }$

Putting the values, ∝ = $\sqrt{\frac{1.8 * 10^{-5} }{0.01} }$

                                    = $\sqrt{18*10^{-4} }$

                                   = 4.24 × 10⁻²

Answer 2

We have Given :

Dissociation constant $(K_{a} ) = 1.8 * 10^{-5}$

Concentration (c) $= 0.01 M$

We have To find :

Percent dissociation

Use the Following Formula:

1. $K_{a} = a^{2} c$

2. Percent dissociation $= \alpha * 100$

Lets Calculation :

$c = 0.01 M = 1 * 10^{-2} M$

Now,

using the formula,

$\therefore \alpha \sqrt{\frac{K_{a} } {c} }$

$=\sqrt{\frac{1.8*10^{-5} }{1*10^{-2} } } =\sqrt{1.8*10^{-3} } =\sqrt{1.8*10^{-4} }$

$=4.242*10^{-2}$

Further, we solve using the formula,

Percent dissociation $= \alpha * 100 = 4.242 * 102 *100 = 4.242\%$

Percent dissociation of $0.01 M$ the acetic acid solution is $4.242\%.$