How you would prepare a 500 mL solution of 0.1M "tris" buffer at pH of 7.5...

How you would prepare a 500 mL solution of 0.1M "tris" buffer at pH of 7.5 starting with solid tris free base. The free base has a chemical forumla of (HOCH2)3-C-NH2 and a MW of 121.1 g/mol. The protonated amine form, tris-chloride, has a pKa of 8.10. To prepare the solution, 1.0M HCl will be used.

Expert Solution

pH = pka + log [ conjugate base] / [ acid] where TrisH+ is acid and Tris is conjugate base

We get 7.5 = 8.1 + log [ Tris] /[TrisH+}

[Tris] = 0.25 [ TrisH+] ......(1)

we were given buffer concentration = 0.1 M . Hence we get

[Tris] + [TrisH+] = 0.1 ..........(2)

solving (1) and ( 2) we get [TrisH+] = 0.08 M , [Tris] = 0.02 M

volume of solution = 500 ml = 0.5 L ,

Moles of TrisH+ = M x V = 0.08 x 0.5 = 0.04 mol , Tris moles = 0.02 x 0.5 = 0.01 mol

we have equation Tris (aq) + H+ (aq) <---> TrisH+ (aq)

Hence we initially need 0.04+0.01 = 0.05 moles of Tris . Later 0.04 mole sof Tris is converted to TrisH+ by addition of 0.04 mol H+ ( i.e HCl) and thus we are left with 0.01 mol Tris

Mass of Tris initially needed = moles of Tris x molar mass of Tris = 0.05 mol x 121.1 g/mol = 6 g

HCl moles needed = Tris H+ moles formed = 0.04mol , Mlarity of HCl = 1M , we find volume

Molarity = moles / volume. Hence we get 1mol/L = 0.04mol / volume of HCl

HCl volume = 0.04/1 = 0.04 L = 40 ml

Thus by taking 6 g of TRIS and adding 40 ml of HCl and then dilutig solution until 500 ml mark with water we get required solution 0.1 M TRIS buffer with pH= 7.5

queen_honey_blossom answered 2 years ago

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