Question

The ionization constant of dimethylamine is 5.4 *10 ^-2 .Calculate its degree of dissociation in its 0.02 M solution .What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Answer 1

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Answer 2

Correct Question:

The ionization constant of dimethylamine is  5.4 × $10^{-4}$ . Calculate its degree of dissociation in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Answer:

The percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH is 0.54 %

Explanation:

Given Data:

The ionization constant of dimethylamine ( $k_{b}$ ) = 5.4 × $10^{-4}$

Concentration (c) = 0.02 M

To Calculate :

If the solution contains 0.1 M NaOH as well, the proportion of dimethylamine that is ionized.

Calculations:

The degree of dissociation is given by

$y = \sqrt{\frac{k_{b} }{c} }$

Substituting the values, we have

$y = \sqrt{\frac{k5.4 \times 10^{-4} }{2 \times 10^{-2} } }$ = 0.164

Let the amount of dimethyl aniline that has been dissociated in the presence of 0.1 M NaOH is denoted by x.

$(CH_{3} )_{2}NH + H_{2} O$ ⇄ $[(CH_{3} )_{2}NH]^{+} + OH^{-}$

The equilibrium concentrations are:

$[(CH_{3} )_{2}NH]$ = 0.02 - x ≈ 0.02

$[(CH_{3} )_{2}NH]^{+} =$ x

$[ OH^{-}]$ = 0.1 + x ≈ 0.1

The equilibrium constant expression is:

$k_{b} = \frac{[[(CH_{3} )_{2}NH]^{+}] + [OH^{-}] }{[(CH_{3} )_{2}NH]} = \frac{x \times 0.1}{0.02}$

So, x = $1.8 \times 10^{-4}$

The percentage of dimethylamine ionized is $\frac{1.8 \times 10^{-4} }{0.02} \times 100= 0.54$

Therefore,  the percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH is 0.54 %

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