Calculate the ratio of total aspirin concentration (#"HAsp" + "Asp"^(-)#) on the L/R side of a membrane between blood plasma and the stomach, if #["HAsp"]# is the same on each side of the membrane? The stomach #"pH"# is #1.00# and blood #"pH"# is #7.4#.

All we have to do here is

1. Calculate the #"HAsp/Asp"^(-)# ratio on each side of the membrane.
2. Note that #["HAsp"]# is the same across the membrane by assuming the blood plasma and stomach are in equilibrium.
3. That allows for cancellation of #["Asp"^(-)]_(L//R)# upon knowing the only other unknown.

And hopefully we recognize we should invoke the Henderson-Hasselbalch equation:

#"pH" = "pK"_a + log\frac(["Asp"^(-)])(["HAsp"])#

Here we distinguish between the left and right sides of the membrane, i.e.

#"pH"_L = "pK"_a + log\frac(["Asp"^(-)]_L)(["HAsp"]_L)#

#"pH"_R = "pK"_a + log\frac(["Asp"^(-)]_R)(["HAsp"]_R)#

Since the protonated aspirin is in equilibrium across the two sides of the membrane, #["HAsp"]_L = ["HAsp"]_R -= ["HAsp"]#. Therefore, and plugging in the #"pH"# and #"pK"_a# values:

#7.4 = 3.5 + log\frac(["Asp"^(-)]_L)(["HAsp"])#

#1.0 = 3.5 + log\frac(["Asp"^(-)]_R)(["HAsp"])#

And so we solve for the two ratios:

#(["Asp"^(-)]_L)/(["HAsp"]) = 10^(7.4 - 3.5) = 7.943 xx 10^3#

#(["Asp"^(-)]_R)/(["HAsp"]) = 10^(1.0 - 3.5) = 3.162 xx 10^(-3)#

Now, since the concentrations of protonated aspirin are equal on either side of the membrane, we obtain our key relationship:

#["HAsp"] = (["Asp"^(-)]_L)/(7.943 xx 10^3) = (["Asp"^(-)]_R)/(3.162 xx 10^(-3))#

Now we have enough info to get the ratio of total drug on either side, i.e. what is

#(["HAsp"] + ["Asp"^(-)]_L)/(["HAsp"] + ["Asp"^(-)]_R) ?#

That can be done as follows.

#(["HAsp"] + ["Asp"^(-)]_L)/(["HAsp"] + ["Asp"^(-)]_R)#

#= ((["Asp"^(-)]_L)/(7.943 xx 10^3) + ["Asp"^(-)]_L)/((["Asp"^(-)]_R)/(3.162 xx 10^(-3)) + ["Asp"^(-)]_R)#

#= ((["Asp"^(-)]_L)/(7.943 xx 10^3) + ["Asp"^(-)]_L)/((["Asp"^(-)]_L)/(7.943 xx 10^3) + ["Asp"^(-)]_R)#

#= ((["Asp"^(-)]_L)/(7.943 xx 10^3) + ["Asp"^(-)]_L)/((["Asp"^(-)]_L)/(7.943 xx 10^3) + ["Asp"^(-)]_L cdot (3.162 xx 10^(-3))/(7.943 xx 10^3))#

where we have used the final relation shown above to substitute for #(["Asp"^(-)]_R)/(3.162 xx 10^(-3))# and #["Asp"^(-)]_R#.

We obtain:

#color(blue)((["HAsp"] + ["Asp"^(-)]_L)/(["HAsp"] + ["Asp"^(-)]_R))#

#= ((cancel(["Asp"^(-)]_L))/(7.943 xx 10^3) + cancel(["Asp"^(-)]_L))/((cancel(["Asp"^(-)]_L))/(7.943 xx 10^3) + cancel(["Asp"^(-)]_L) cdot (3.162 xx 10^(-3))/(7.943 xx 10^3))#

#= (1/(7.943 xx 10^3) + 1)/(1/(7.943 xx 10^3) + (3.162 xx 10^(-3))/(7.943 xx 10^3))#

#= (1 + 7.943 xx 10^3)/(1 + 3.162 xx 10^(-3))#

#= 7919.24#

#= color(blue)(7.92 xx 10^3)#