What is the pOH of a #0.50 M# solution of acetic acid? The acid dissociation constant of acetic acid at #25.0^@ "C"# is #1.75 xx 10^-5#.

1 Answer
Apr 14, 2018

#pOH=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?