- #1
Castiel
- 3
- 0
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?
Last edited: