Henderson-Hasselbalch Calculations
for 500 ml of 0.5 M Tris-Cl, pH 6.8
- Mathematical Proof -
| Equilibrium | Description | Components |
|---|---|---|
| HA ⇌ H⁺ + A⁻ | HA = Undissociated Acid A⁻ = Conjugate Base of HA H⁺ = Hydrogen Ion |
| Ka = [H⁺][A⁻]/[HA] | Therefore: -log[H⁺] = -log[Ka] + log[A⁻] - log[HA] |
| pKa = -log[Ka] | pH = -log[H⁺] |
| log[A⁻] - log[HA] = log[A⁻/HA] |
Henderson-Hasselbalch Equation: pH = pKa + log[A⁻/HA]
Tris-base : MW = 121.14 ; pKa = 8.12 at 25°C
Since the desired pH is 6.8, then 6.8 = 8.12 + log[A⁻/HA]
Therefore, log[A⁻/HA] = -1.320
∴ [A⁻/HA] = 10(-1.320) ≈ 0.048
The concentration of conjugate base [A⁻] is equal to the concentration of base (in this case 0.5 M) minus the concentration of H⁺ (which is what we are solving for). Similarly, the concentration of the undissociated acid is equal to the concentration of H⁺. Thus, [A⁻/HA] = ([Base] - [Acid] ) / [Acid] = 0.048
Multiply both sides of the equation by [Acid] to get : [Base] - [Acid] = (0.048) * [Acid]
Add [Acid] to both sides of the equation to get : [Base] = ((0.048) * [Acid] )+ [Acid]
Divide both sides of the equation by [Acid] to get : [Base]/[Acid] = (0.048) + 1
Invert the equation and get : [Acid] = [Base] / (0.048 + 1)
∴ [Acid] = [0.5 M] / (0.048 + 1) ≈ 0.4771 M
If 12.1 M Hydrochloric Acid is used then [12.1] / [0.4771] ≈ 25.36 (dilution factor)
Required Volumes:
Divide the desired volume (500 ml) by the dilution factor to get the required volume of HCl to add:
500 ml / 25.36 ≈ 19.72 mL of 12.1 M HCl needed
To determine the amount of Tris-base needed, multiply the desired volume (500 ml), by the desired concentration (0.5 M), by the molecular weight (121.14) :
(0.5 L) x (0.5 M) x (121.14 g/mol) = 30.285 g