Question

1. Tris buffer

a. 100 mM Tris-HCl, pH 8.0

b. 100 mM Tris-HCl, pH 6.0

Make this buffer by dissolving an appropriate amount (which you must calculate using the formula weight) of solid Tris in APPROXIMATELY 80 mL of water. Use the Henderson-Hasselbalch equation to calculate how much 1 M HCl is needed to achieve the final pH?

Answer 1

pKa of tris-HCl = 8.08

(a): Cncentration of Tris-HCl buffer need to be prepared = 100 mM = 100 mM x (1M / 1000 mM) = 0.100 M

Volume of Tris-HCl buffer need to be prepared = 80 mL = 0.080 L

Hence moles of Tris need to be added = MxV = 0.100 mol/L x 0.080 L = 0.008 mol

mass of Tris need to be taken =  0.008 mol x 121.14 g/mol = 0.969 g

Applying Hendersen equation:

pH = pKa + log([Tris] / [Tris-HCl])

=> 8.0 = 8.08 + log([Tris] / [Tris-HCl])

=> [Tris] / [Tris-HCl] = 10^-0.08 = 0.832

=> (moles of Tris) / (moles of Tris-HCl) = 0.832

=> (0.008 mol - y) / y = 0.832

=> (0.008 mol - y) = 0.832y

=> 0.008 mol = 1.832y

=> y = 0.00437 mol

Hence moles of Tris-HCl need to be made = 0.00437 mol

=> moles of HCl need to be added = 0.00437 mol

Concentration of HCl = 1M

=> MxV = 0.00437 mol

=> 1 mol/L x V = 0.00437

=> V = 0.00437 L = 4.37 mL (answer)

(b): pH = 6.0

Applying Hendersen equation:

pH = pKa + log([Tris] / [Tris-HCl])

=> 6.0 = 8.08 + log([Tris] / [Tris-HCl])

=> [Tris] / [Tris-HCl] = 10^-2.08 = 0.00832

=> (moles of Tris) / (moles of Tris-HCl) = 0.00832

=> (0.008 mol - y) / y = 0.00832

=> (0.008 mol - y) = 0.00832y

=> 0.008 mol = 1.00832y

=> y = 0.00793 mol

Hence moles of Tris-HCl need to be made = 0.00793 mol

=> moles of HCl need to be added = 0.00793 mol

Concentration of HCl = 1M

=> MxV = 0.00793 mol

=> 1 mol/L x V = 0.00793

=> V = 0.00793 L = 7.93 mL (answer)