Question #9a945

1 Answer
Jun 21, 2017

Answer:

We need #pK_a# or #K_a# for acetic acid. And under other circumstances I might make you dig for it..........

Explanation:

This site tells me that #pK_a(HOAc)=4.76#. Acetic acid is thus a fairly weak acid.......

And now we interrogate the equilibrium......

#HOAc(aq) + H_2O(l) rightleftharpoons H_3O^(+) + AcO^(-)#....

And thus we can write.......

#([H_3O^+][AcO^-])/([HOAc])=10^(-4.76)#

We know that, initially, #[HOAc]=0.540*mol*L^-1#, and we guess that #x*mol*L^-1# of acetic acid undergoes protonolysis...

And thus #10^(-4.76)=(x xx x)/(0.540-x)=x^2/(0.540-x)#

This is a quadratic in #x#, which we could solve exactly. Because chemists are lazy folk, however, we ASSUME that #0.540">>"x#, and thus #0.540-x~=0.540#

And so #x_1=sqrt(10^(-4.76)xx0.540)=0.00306#

We can recycle this expression, i.e. we use successive approximations, to check on how good it was with respect to the initial assumptions.......

#x_2=sqrt(10^(-4.76)xx(0.540-0.00306))=0.00305#

And this value has converged....so the approximation is good.

Now by definition of our problem #x=[H_3O^+]=0.00305#, and #pH=2.52#, and thus #pOH=11.48#.

And finally, #[HO^-]=10^(-11.48)=3.27xx10^-12*mol*L^-1#.