If you mixed equal volume of .1 M HCl and .20 M Tris (pka=8.3), IS THE RESULTING SOLUTION A BUFFER SOLUTION?

Answer 1

Work it out to see. Take 100 mL of each, then
millimols tris = 100 x 0.2 = 20
mmols HCl = 100 x 0.1 = 10
Here is what you have initially.
..................XNH2 + HCl ==> XNH3^+ + Cl^-
Initial...........20...........0..............0
add.........................10....................
Change.....-10........-10............+10
Equil...........10..........0...............10

pH = pKa + log (base/acid)
pH = 8.3 + log [(10/200)/(10/200)]
pH = 8.3 + log 1 = 8.3

So you have XNH2 and the salt XNH3^+ which makes a buffer. Try it by adding say 10 mL of 0.1 M HCl OR 10 mL of 0.1 M NH3. Here is how the addition of HCl would look. mmol HCl added = 10 x 0.1 = 1
.................XNH2 + HCl ==> XNH3^+ + Cl^-
I.....................10.......0............10..............
add...........................1..............................
C.....................-1......-1............+1
E......................9........0.............11
Then pH = 8.3 + log (b/a)
pH = 8.3 + log [(9/210)/(11/210)] = ?
pH = about 8.2, Adding HCl should make the buffer more acid and it does. You can do the NH3 one. Hope this helps.