Question

8. Tris buffers tris(hydroxymethyl)aminomethane) are extensively used in biochemistry. The pK, for Tris is 8.1. Mixing Tris base (A', f.w. 121.1) with Tris-HCI (HA, f.w. 157.56) provides buffering in the pH range of 7 to 9. How many grams of the acid form and base form need to be dissloved in a total volume of 0.5 L to obtain a 0.25 M buffer solution at pH 8.8?

Answer 1

Sol.

Let the moles  of Tris - HCl and Tris base be x and y respectively

As Volume of buffer = 0.5 L

Conc. of buffer = 0.25 M  

So , Total moles of buffer = 0.25 × 0.5 = 0.125 mol

and , x + y = 0.125  

Now , pH = 8.8 and pKa = 8.1

Using Henderson - Hasselbalch equation ,

pH = pKa + log ( moles of Tris base / moles of Tris - HCl )

8.8 = 8.1 + log ( y / x )

log ( y / x ) = 0.7

y / x = 10^0.7 = 5.0118  

y = 5.0118 x

So , x + y = x + 5.0118 x = 0.125

6.0118 x = 0.125  

x = 0.125 / 6.0118 = 0.0208 mol

and y = 5.0118 x = 5.0118 × 0.0208 = 0.1042 mol  

As molar mass of Tris - HCl = 157.56 g / mol

mass of Tris - HCl = 0.0208 × 157.56 =   3.2772 g  

molar mass of Tris base = 121.1 g / mol

mass of Tris base = 0.1042 × 121.1 =   12.6186 g