What is the pOH of a 0.50 M0.50M solution of acetic acid? The acid dissociation constant of acetic acid at 25.0^@ "C"25.0C is 1.75 xx 10^-51.75×105.

1 Answer
Apr 14, 2018

pOH=11.47pOH=11.47

Explanation:

We address the equilibrium....

underbrace(HOAc(aq))_"acetic acid" + H_2O(l)rightleftharpoons""^(-)OAc +H_3O^+

And of course, while this will give us H_3O^+, and thus pH, pOH=14-pH...now given the equilibrium expression, and if we put [""^(-)OAc]=[H_3O^+]=x...then....

K_a=([AcO^(-)][H_3O^+])/([HOAc])=x^2/(0.50*mol*L^-1-x)=1.75xx10^-5.

This is a quadratic in x that COULD be solved EXACTLY....but chemists are lazy....and so we make the approximation that 0.50">>"x...and thus 0.50-x~=x....you know the usual story, approximate, then justify, then re-approximate using the first approximation....

And so x_1=sqrt{0.50xx1.75xx10^-5}=2.96xx10^-3*mol*L^-1...the which value is indeed small compared to 0.50*mol*L^-1...but given an approximation for x, we resubstitute this back into the expression...

x_2=sqrt{(0.50-2.96xx10^-3)xx1.75xx10^-5}=2.95xx10^-3*mol*L^-1

x_3=sqrt{(0.50-2.95xx10^-3)xx1.75xx10^-5}=2.95xx10^-3*mol*L^-1

...and since the approximations have converged, I am prepared to accept this as the TRUE value...the same as if we solved the quadratic equation...

But x=[H_3O^+]=[""^(-)OAc] by our definition....and [HOAc]=0.50-2.95xx10^-3)*mol*L^-1=0.497*mol*L^-1...

And so [H_3O^+]=[""^(-)OAc]=2.95xx10^-3*mol*L^-1...

pH=-log_10{2.95xx10^-3}=-(-2.53)=2.53

And since...

K_w=[H_3O^+][HO^-]=10^(-14)...

And we can take log_10 of both sides to give....

log_(10)K_w=log_(10)10^(-14)=log_10[H_3O^+]+log_10[HO^-].

And thus.... -14=log_(10)[H_3O^+]+log_(10)[HO^-]

Or.....

14=-log_(10)[H_3O^+]-log_(10)[HO^-]

14=underbrace(-log_10[H^+])_(pH)underbrace(-log_10[OH^-])_(pOH)

14=pH+pOH
By definition, -log_10[H^+]=pH, -log_10[HO^-]=pOH

pOH=14-pH=14-2.53=11.47...for the given solution under standard conditions.....got all that?