Question

A solution contains 32 g/L of tris(hydroxymethyl)-aminomethane, also known as TRIS base (pKb= 5.9) (Molec Wt= 121.12). to 35 mL of this TRIS solution, 25 ml of 0.1N HCl solution is added and the volume is made up to 100 mL. What is the pH of the solution? It is known that one mole of TRIS base reacts with one mole of HCL to give one mole of TRIS-HCL salt

Answer 1

For TRIS solution,

1 L contains 32 g of TRIS.

So, 35 ml of this solution would contain TRIS = 32*0.035 = 1.12 g

Molar mass of TRIS = 121.12 g/mol

Moles of TRIS in 35 ml = 1.12/121.12 = 0.0092 mol

Moles of HCl = concentration*volume = 0.1*0.025 = 0.0025 mol

Moles of HCl < Moles of TRIS

TRIS + HCl ----> TRIS-HCl

So, whole of HCl has reacted.

Moles of TRIS remaining = 0.0094 - 0.0025 = 0.0069 mol

Total volume = 100 ml = 0.100 L

Concentration of TRIS = 0.0069/0.100 = 0.069 M

SO, TRIS will be determing pH of the solution.

TRIS is a weak base, so to determine pH of the solution,

TRIS + H2O ----> TRIS-H+ + OH-

	TRIS	TRIS-H+	OH-
Initial	0.069 M	0	0
Change	-x	+x	+x
Equilibrium	0.069-x	x	x

pKb = 5.9

Kb = 10^-5.9 = 1.26*10^-6

Kb = 1.26*10^-6 = x²/0.069

x = 0.00069

x = [OH-] = 0.00069

pOH = -log (0.00069) = 3.5

pH = 14-3.5 = 10.5

pH = 10.5