Finding Ka using a Galvanic cell and Electric Potential

[Castiel]

Homework Statement

Suppose you wish to determine Ka for a monoprotic weak acid of molecular mass 76.11 g/mol. The pH meter you need to use is not working. You decide to weigh out 2.28 g of the acid and 1.96 of its sodium salt and dissolve them together in water, add quinhydrone and set up a half-cell with Pt wire just as you did in this experiment. Unfortunately, the only other wire you can find is aluminum. Therefore, you prepare a .1 M solution of aluminum sulfate, create an Al3+/Al half-cell and set up a galvanic cell with the quinhydrone electrode. The standard reduction potential for the aluminum couple is -1.67 V. When you measure the cell potential, you find it to be 2.20V. From this information, calculate Ka for the weak acid.

Homework Equations

Ecell=Estd-(0.0257/n)*ln(Ka) ?
2.20v=(-1.67v+E quinhydrone cell)-(0.0257/3)*ln(Ka)

Al3+ + 3e- → Al (s) Eo=-1.67v
quinone+ 2H+ +2e- → hydroquinone ?

The Attempt at a Solution

For the E of the quinhydrone cell, would I be able to use this (given earlier in the lab)?

quinone+ 2H+ +2e- → hydroquinone Eo=.70v

And I have no idea what to do with the grams or the molar mass. All I can think of is that you could convert it into moles and somehow find the molarity from that? Assuming it's one liter? But can you do that? Can you just assume it's one liter?

And what is the "1.96 g of its sodium salt" mean? is that the aluminum sulfate?

 

[Borek]

Sodium salt of the acid. Your solution contains both acid and its salt - something like mixture of acetic acid and sodium acetate. Think in terms of Henderson-Hasselbalch equation.

Nice question.

 

[Castiel]

Hmmm so I'd use this equation?

[H+]=Ka [HA]/[A-] ?
Then to find the concentrations, could I just do...
(2.28g H+) X (1 mol/76.11 g) = .02996 mol H+ and for one liter-----> [H+]=0.02996 and
(1.96g A-) X (1 mol/76.11 g) = .02575 mol A- and so [A-]=0.02575

meaning...
0.02996=Ka [HA]/.02575 ?
But how do I find HA? I'm assuming it has something to do with the electric potential...would I use .70 for my volts for the acid half cell?
Then I know the concentration of the aluminum sulfate...and I have the acid and base concentrations...so I could just plug those into the Ecell equation and solve for [HA] then plug that into the [H+]=Ka [HA]\[A-] equation and solve for Ka?

Another question. Would the aluminum half cell be the cathode?

 

[Borek]

IMHO you are expected to use potential to find pH, HA and A^- are known - you are told what masses were dissolved, so you can calculate molarities.