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
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
= (6,4 × 10^-5) / 0.01
= 6.4 × 10^-3
Degree of dissociation of (C₂H₅)₃N is 6.4 × 10^-3
Answered by
113
The degree of dissociation is
Given:
C = 0.1 M
Solution:
The formula to calculate the degree of association is as follows:
We assume that is very small and thus apply the direct formula.
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
We know that,
Therefore,
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