For a solution that is 0.285 M #HC_3H_5O_2# (propionic acid, #K_a=1.3×10−5#) and 0.0888 M HI, calculate the following?

1 Answer
Mar 17, 2017

Answer:

We interrogate the equilibrium:

#"H"_3"CCH"_2"CO"_2"H"+"H"_2"O"rarr"H"_3"CCH"_2"CO"_2^(-)+"H"_3"O"^+#

Explanation:

#"H"_3"CCH"_2"CO"_2"H"+"H"_2"O"rarr"H"_3"CCH"_2"CO"_2^(-)+"H"_3"O"^+#

And this equilibrium is governed by a constant, #K_a#, where.......

#K_a=(["H"_3"CCH"_2"CO"_2^(-)]["H"_3"O"^+])/(["H"_3"CCH"_2"CO"_2"H"])=1.3xx10^-5#

But the #["H"_3"O"^+]# will be artificially high, because #"HI"# has been added to solution, and its dissociation will be reasonably complete. Now we make the approximation that #x*mol*L^-1# of #"propionic acid"# dissociates, and thus we reformulate the equilibrium expression as follows:

#K_a=((x)(0.0888+x))/(0.285-x)=1.3xx10^-5#

And so we solve for #x#, by assuming that #0.0888>x#, and #0.285>x#. This approx. may not be justified:

#x_1~=(1.3xx10^-5xx0.285)/(0.0888)=4.2xx10^-5#

Given this approx. we can recycle it back into the equilibrium expression, and see how #x# evolves........

#x_1~=(1.3xx10^-5xx0.285)/(0.0888)=4.2xx10^-5#

#x_2=4.17xx10^-5#. Since the values of #x# do not change substantially, our approx., was correct and we can continue with this estimate.

And thus #[""^(-)"O(O=)CCH"_2"CH"_3]=4.2xx10^-5*mol*L^-1#

#["I"^-]=0.0888*mol*L^-1#, because we have reasonably assumed that all the #"hydroiodic acid"# has dissociated..........