Question

Calculate the volume, in milliliters, of a 0.330 M KOH solution that should be added to 6.000 g of HEPES (MW = 238.306 g/mol, pKa = 7.56 ) to give a pH of 7.36.

Answer 1

Answer:

$\large \boxed{\text{ 29.3 mL}}$

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

$n = \text{6.00 g} \times \dfrac{\text{1 mol}}{\text{238.306 g}} = \text{0.025 18 mol}$

2. Calculate the concentration ratio.

We can use the Henderson-Hasselbalch equation.

$\begin{array}{rcl}\text{pH} &= &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\7.36 & = & 7.56 + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\-0.20 & = & \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\dfrac{[\text{A}^{-}]}{\text{[HA]}} & = & 0.631\\\\\end{array}$

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⁻

$\begin{array}{rcl}\text{A}^{-} & = & \text{0.631 HA}\\\text{A ^{-} + HA} & = & 0.02518\\\text{0.631 HA + HA} & = & 0.02518\\\text{1.631 HA} & = & 0.02518\\\text{HA} & = & 0.015 44\\\end{array}$

$\text{A ^{-} } = 0.009 742$

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

$V = \text{0.009 742 mol KOH} \times \dfrac{\text{1 L KOH}}{\text{0.330 mol KOH}} = \text{0.0293 L KOH} = \text{29.3 mL KOH}\\\\\text{We must add \large \boxed{\textbf{ 29.3 mL}} of the KOH solution.}$