Final answer:
To prepare a pH 7.79 Tris buffer with 31.52 g of TrisHCl, we calculated that we would need 0.00978 L (9.78 mL) of 10.0 M NaOH using the Henderson-Hasselbalch equation, option c is correct.
Step-by-step explanation:
To calculate the volume of 10.0 mol L⁻¹ NaOH needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
First, we find the moles of TrisHCl using its molar mass. The molar mass of TrisHCl is approximately 157.6 g/mol. So, the moles of TrisHCl in 31.52 g are:
moles of TrisHCl = 31.52 g / 157.6 g/mol = 0.2 moles
Assuming Tris has a pKa of 8.1, we can rearrange the Henderson-Hasselbalch equation to solve for [A-], which is the concentration of the conjugate base (Tris):
[A-] = [HA] * 10^(pH - pKa)
[A-] = 0.2 moles * 10^(7.79 - 8.1)
[A-] = 0.2 moles * 10^(-0.31) = 0.2 moles * 0.489 = 0.0978 moles
The volume of NaOH needed can then be calculated using the molarity and the number of moles of [A-]:
Volume = moles / molarity
Volume = 0.0978 moles / 10.0 mol/L = 0.00978 L = 9.78 mL
None of the provided options (a) 2.56 L, (b) 3.78 L, (c) 4.91 L, (d) 6.23 L match the calculated volume; therefore, there might be a mistake in the available options or in the question itself.