Question

What is the pH of a 0.050 M triethylamine (C2H5)3N solution? (Kb= 5.3 x 10^-4)

Please include steps for how to solve the problem. Thank you! :)

Answer 1

`pH=11.7`

Explanation:

We address the equilibrium...

`H_2O(l) + N(CH_2CH_3)_3(aq) rightleftharpoonsHO^(-) + Hstackrel+N(CH_2CH_3)_3`

For which `K_b=([HO^(-)][Hstackrel+N(CH_2CH_3)_3])/([N(CH_2CH_3)_3(aq)])`...

And now we simply put in some numbers, and NOTE that `[HO^-]=[Hstackrel+N(CH_2CH_3)_3]=x`...so...

`K_b=x^2/(0.050-x)=5.3xx10^-4`...and if `0.050">>"x`...then...

`x_1~=sqrt(5.3xx10^-4xx0.05)=4.6xx10^-3`...and now we got a first approx. for `x`...and we can run with this...

`x_2~=sqrt(5.3xx10^-4xx(0.05-4.6xx10^-3))=4.91xx10^-3`

`x_3~=sqrt(5.3xx10^-4xx(0.05-4.91xx10^-3))=4.89xx10^-3`

`x_4~=sqrt(5.3xx10^-4xx(0.05-4.89xx10^-3))=4.89xx10^-3`

You notice that the approximations get betterer and betterer the more we make and I am prepared to accept `x=4.89xx10^-3*mol*L^-1=[HO^-]` etc...

And so `pOH=-log_10(4.89xx10^-3)=2.31`

`pH=14-2.31=11.7`........does you follow?

Alternatively this could solved by the quadratic equation. The approximation method is generally a bit easier than this approach. It is up to you to be able to use BOTH methods...