Chemistry, asked by amanjoth994, 1 year ago

The ionisation constant of base (C2H5)3N is 6.4*x10^-5 Calculate its degree of dissociation in its 0.1M solution when it is mixed with 0.01 M NaOH solution

Answers

Answered by JunaidMirza
224
Degree of dissociation = Ka / Concentration of strong electrolyte having common ion added
= (6,4 × 10^-5) / 0.01
= 6.4 × 10^-3

Degree of dissociation of (C₂H₅)₃N is 6.4 × 10^-3
Answered by skyfall63
113

The degree of dissociation is 6.4 \times 10^{-3}

Given:

\mathrm{K}_{\mathrm{a}}=6.4 \times 10^{-5}

C = 0.1 M

Solution:

The formula to calculate the degree of association is as follows:

\alpha=\sqrt{\frac{K_a}{C}}

We assume that   is very small and thus apply the direct formula.

\alpha=\sqrt{\frac{6.4 \times 10^{-5}}{0.1}}=2.53 \times 10^{-2}

In the presence of NaOH, the equilibrium will shift backward and the concentration of base will increase.

Let c be the be the initial concentration and x the amount now dissociated, then at equilibrium

\left[\left(C_{2} H_{5}\right)_{3} N\right]=c-x

\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NH}^{+}=\mathrm{x}

\left[\mathrm{OH}^{-}\right]=0.01+\mathrm{x}

\therefore \mathrm{K}_{\mathrm{a}}=\frac{x(0.01+x)}{c-x}=\frac{x \times 0.01}{c}

\frac{K_a}{0.01}=\frac{x}{c}

We know that,

\alpha=\frac{x}{c}

Therefore,

\alpha=\frac{K a}{0.01}=\frac{6.4 \times 10^{-5}}{0.01}=6.4 \times 10^{-3}

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