Question

How many mL of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride to give a pH of 7.40 in a final volume of 500 mL?

Short answer

The volume of sodium hydroxide required to form buffer solution is 22.3mL

Step-by-step solution

Define the Henderson-Hasselbalch Equation.

The relationship between acid pH and pKa in aqueous solutions is described by the Henderson-Hasselbalch equation (acid dissociation constant). This equation can be used to forecast the pH of a buffer solution when the concentrations of such acid as well as its conjugate base, or the base and the related conjugate acid, are determined.

Calculate the volume of sodium chloride.

$+{\mathrm{OH}}^{-}\to \mathrm{Tris}$The amount of sodium hydroxide that is needed to make a buffer solution.

Tris(hydroxymethyl)-aminomethane is written as tris for our convenience.

Tris hydrochloride has a mole count of

$=\frac{10.0\mathrm{g}}{157.596\mathrm{g}/\mathrm{mol}}=0.0637\mathrm{mol}$

Tris hydrochloride is made by adding sodium hydroxide to a tris hydrochloride buffer solution.

Tris

$+{\mathrm{OH}}^{-}\to \mathrm{Tris}$

Using the Henderson-Hasselbalch equation,

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\mathrm{I}\mathrm{tris}}{\left[\mathrm{trishydrochloride}\right]} \\ 7.40=8.072+\log \frac{\mathrm{x}}{0.0635}-0.672 \\ =\log \frac{\mathrm{x}}{0.0635-\mathrm{x}} \end{gathered}$$

By dividing log onboth sides we get

$$\begin{gathered} {10}^{-0.672}=\frac{x}{0.0635-\mathrm{x}} \\ x=0.0111 \end{gathered}$$

The volume of sodium hydroxide in 0.0111 mol is

$=\frac{0.0111\mathrm{mol}}{0.500\mathrm{mol}/\mathrm{L}}=22.3\mathrm{mL}$

Thus, the volume of sodium hydroxide required to form a buffer solution 22.3mL