[Win,odd Dhamnekar]

What would the % ionization be for a 0.21 molar solution of Piperidine(K=1.3× 10^-3)?

Since Piperidine is a weak base, [OH^-] is determined using its K_b of 1.30×10^-3.

Using the standard approaches yields a value of 0.0158

% ionization=[OH^-]/C_i=0.0158/0.21*100=7.55

Below are listed the values for the seven variables in such a solution.

pH=12.2
pOH=1.80
[H+]=6.31×10^-13
[OH-]=0.0158
% ionization=7.55
[C5H11N]=0.194
[C5H12N⁺]=0.0158



In my opinion, above answers looks correct. What is your opinion?

[Orcio_87]

Using the standard approaches yields a value of 0.0158

% ionization=[OH-]/Ci=0.0158/0.21*100=7.55

I calculated it twice and - if ionisation degree = √(K/C) - I still get that it will be 7,87 %.

But - since ionisation is bigger than 5 % I think it should be calculated without shortcuts (C - αC ≠ C). I calculated that ionisation degree should be around 3 %.

[mjc123]

I agree with OP's answer. Doing it the more accurate way is never going to make as big a difference as 7% to 3%. The difference is only a factor of 1-α, or ca. 0.93.