Calculate ionic concentration

[Krushnaraj Pandya]

Homework Statement

The ionization constant of ${(C_2 H_5)}_3 {N}$ is 6.4 x 10^(-5). Calculate degree of dissociation in its 0.1 M solution when it is mixed with 0.01 M NaOH solution.

2. The attempt at a solution
the compound gives ${(C_2 H_5)}_3 {NH+}$ and $OH-$. C is initial concentration of compound=0.1 M and x as degree of dissociation- C*x is final concentration of ${(C_2 H_5)}_3 {NH+}$ and since NaOH is strong electrolyte 0.01 M OH- is present. Since x is small compared to 0.01 we take OH concentration as 0.01. Now (C*x)(0.01)=K. placing 0.1 as C we get x as 6.4 x 10^(-2). However the answer is 6.4 x 10^(-3). Where am I wrong?

 

[Krushnaraj Pandya]

Never mind. I totally forgot to write C in denominator. Silly me.

 

[epenguin]

You are saying you now get the same answer as the book? I got the same answer so the book appears to be right. Question phrasing a bit loose.

 

[Krushnaraj Pandya]

You are saying you now get the same answer as the book? I got the same answer so the book appears to be right. Question phrasing a bit loose.

yes, I did- thank you. I was just making a really silly mistake.