Answer:
[tex]\large \boxed{\text{ 29.3 mL}}[/tex]
Explanation:
HEPES is a zwitterion. That is, it has both the acid and base components in the same molecule. However, we can write its formula as HA. Then the equation for the equilibrium is
MM: 238.306
HA + H₂O ⇌ H₃O⁺ + A⁻; pKₐ = 7.56
m/g: 6.00
1. Calculate the moles of HEPES
[tex]n = \text{6.00 g} \times \dfrac{\text{1 mol}}{\text{238.306 g}} = \text{0.025 18 mol}[/tex]
2. Calculate the concentration ratio.
We can use the Henderson-Hasselbalch equation.
[tex]\begin{array}{rcl}\text{pH} &= &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\7.36 & = & 7.56 + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\-0.20 & = & \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\dfrac{[\text{A}^{-}]}{\text{[HA]}} & = & 0.631\\\\\end{array}[/tex]
The acid and its conjugate base are in the same solution, so the concentration ratio is the same as the mole ratio.
3. Calculate the moles of HA and A⁻
[tex]\begin{array}{rcl}\text{A}^{-} & = & \text{0.631 HA}\\\text{A$^{-}$ + HA} & = & 0.02518\\\text{0.631 HA + HA} & = & 0.02518\\\text{1.631 HA} & = & 0.02518\\\text{HA} & = & 0.015 44\\\end{array}[/tex]
[tex]\text{A$^{-}$} = 0.009 742[/tex]
4. Calculate the moles of KOH
We are preparing the buffer by adding KOH to convert HA to A⁻. The equation is
HA + OH⁻ ⟶ A⁻ + H₂O
The molar ratio is 1 mol A⁻:1 mol OH⁻, so we must use 0.009 742 mol of KOH.
5. Calculate the volume of KOH
[tex]V = \text{0.009 742 mol KOH} \times \dfrac{\text{1 L KOH}}{\text{0.330 mol KOH}} = \text{0.0293 L KOH} = \text{29.3 mL KOH}\\\\\text{We must add $\large \boxed{\textbf{ 29.3 mL}}$ of the KOH solution.}[/tex]
