If you mix equal volumes of 0.1 M HCl and 0.2 M TRIS (free amine form), is the r

Question

If you mix equal volumes of 0.1 M HCl and 0.2 M TRIS (free amine form), is the resulting solution a buffer? Why or why not?

The answer is:

Yes, b/c it contains equal concentrations of TRIS in the acid and free amine forms. When the 2 solutions mix, the concentrations (in the absence of a reaction) are 0.05 M HCl and 0.1 M TRIS b/c of dilution. The HCl reacts with half of the TRIS present, giving 0.05 M TRIS (protonated form) and 0.05 M TRIS (free amine form).

BUT I don't understand the answer. Where did the diluted concentrations come from? And then why did it only react with half of the TRIS? Please help !!!

Explanation / Answer

Let us suppose the volume of the two solutions is x

when they are mixed the total volume now is 2x (x+x)

so as we see the HCL = 0.1M

and TRIS = 0.20 M

Molarity = Moles / Volume

so the moles of HCL = 0.1*x = 0.1x

and the moles of TRIS = 0.2*x = 0.2x

now as we see one 0.1 mole HCL will react with 0.1 mole of TRIS and make a salt (protonated form)

therefore the concentration of the protonated form = 0.1x/ total volume

= 0.1x/2x = 0.05 M

and

the remaining 0.1x moles of TRIS will have a concentration = 0.1x/2x = 0.05 M (free amine form )

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