The ionization constant of base ${(C_{2} H_{5})}_{3}$ is $6.4 \times 10^{- 5}$ . Calculate its degree of dissociation in its $0.1 M$ solution when it is mixed with $0.01 M$ $N a O H$ solution.

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Solution

Consider $α$ , be very small ,by applying formula directly.

$α = \sqrt{\frac{K_{a}}{C} = \sqrt{\frac{(6.4 X 1 0^{- 5})}{0.1}}} = 2.53 X 1 0^{- 2}$

In the presence of $N a O H$ , equilibrium will be shift in the backward i.e, the concentration of base will increase.

If C is the initial concentration,

$X$ is the amount dissociated, then equilibrium will be:

[ ${(C}_{2} H_{5})_{3} N$ ] = $c - x,$

${[(C}_{2} H_{5})_{3} N H^{+}$ = $X,$

[O $H^{-}$ ] = $0.01 + X$

$∴ K_{a} = \frac{X (0.01 + X)}{c - x} = \frac{X (0.01)}{X}$

$\frac{K_{a}}{0.01} = \frac{x}{c}$

but $\frac{X}{C}$ means $α$

$α = \frac{K_{a}}{0.01} = \frac{6.4 X 1 0^{- 5}}{0.01} = 6.4 X 1 0^{- 3}$

Hence, $6.4 X 1 0^{- 3}$ is the answer.

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The ionization constant of base (C2H5)3 is 6.4×10−5. Calculate its degree of dissociation in its 0.1 M solution when it is mixed with 0.01 M NaOH solution.

The ionization constant of base ${(C_{2} H_{5})}_{3}$ is $6.4 \times 10^{- 5}$ . Calculate its degree of dissociation in its $0.1 M$ solution when it is mixed with $0.01 M$ $N a O H$ solution.