Question

BUFFER PROBLEMS

You are asked to prepare 1.2 L of a 0.05 M tris buffer at pH= 7.8. You start with the conjugate base form of tris (121 g/mol). How many grams of tris must you weigh out? How many mL of 6 M HCl (a strong acid) must you add to reach pH=7.8? The pKa for tris is 8.1.

Answer 1

Molarity of buffer is 0.05M tris buffer and volume is 1.2 L

Therefore, the mass of tris required = Molarity x mol. Mass x volume in L = 0.05 mol/L x 121 g/mol x 1.2 L = 7.26 g

Hence, we need 7.26 grams of conjugate base tris required to make the 1.2 L of buffer solution.

Now, we need to get to the pH = 7.8;

   tris-acid <---> tris-base + H+ (pka = 8.1)

According to Henderson–Hasselbalch equation:

pH = pKa + log[tris-acid]/[tris-base]

7.8 = 8.1 + log[tris-acid]/[tris-base]

log[tris-acid]/[tris-base]= -0.3

[tris-acid]/[tris-base] = 0.50

We know that [tris-acid]+[tris-base] = 0.05 M

[tris-acid] = 0.5 (0.05 M - [tris-acid])

1.5 [tris-acid] = 0.025
[tris-acid] = 0.0166 M
[tris-base] = 0.0333 M

Therefore, no. moles of tris-acid = molarity x volume in L = 0.0166 mol/L x 1.2 L = 0.01992 mol

tris-base + H+ ---> tris-acid

Hence, from the stochiometry no. moles of tris-acid is equal to no. moles of acid added.

therefore, no. moles of H+ or HCl = 0.01992 mol

The volume of HCl required = V = no. moles / molarity = (0.01992 mol)/(6 moles/L) = 0.00332 L = 3.32 mL

Therefore, we need 3.32 mL of 6M HCl required to reach pH = 7.8.