I have 25 grams of Lidocaine HCl.
Lidocaine have an molar mass of 234.3g/mol and a density of 1.026 g/mL.
HCl have an molar mass of 36.46g/mol and a density of 1.18g/mL
NaOH have an molar mass of 39.997g/mol and a density of 1.515g/mL.
In Lidocaine HCl, there is 1:1 equivalents of Lidocaine and HCl.
How much volume and molarity of NaOH should I add to Lidocaine HCl to convert it to pure Lidocaine base form?
The correct answer and explanation is :
To convert Lidocaine HCl to pure Lidocaine base, a stoichiometric amount of NaOH is required to neutralize the HCl component in the salt. Here’s how to calculate it:
✅ Final Answer (Short Form)
- Volume of 1.0 M NaOH solution required: 0.107 mol × (1 L / 1 mol) = 107 mL
- Molarity of NaOH solution: 1.0 M (assumed for practicality)
🧪 Step-by-Step Explanation
1. Determine moles of Lidocaine HCl
Given:
- Mass of Lidocaine HCl = 25 g
- Molar mass of Lidocaine HCl = Molar mass of Lidocaine + HCl
= 234.3 g/mol + 36.46 g/mol = 270.76 g/mol
So, Moles of Lidocaine HCl=25 g270.76 g/mol≈0.0923 mol\text{Moles of Lidocaine HCl} = \frac{25\ \text{g}}{270.76\ \text{g/mol}} \approx 0.0923\ \text{mol}
But if you’re using 25 g of Lidocaine HCl, this means 0.0923 mol of both Lidocaine and HCl are present (1:1 ratio).
2. Stoichiometry with NaOH
The reaction is: Lidocaine\cdotpHCl+NaOH→Lidocaine (base)+NaCl+H2O\text{Lidocaine·HCl} + \text{NaOH} \rightarrow \text{Lidocaine (base)} + \text{NaCl} + \text{H}_2\text{O}
Thus, 1 mol of NaOH is needed per 1 mol of HCl.
So, moles of NaOH required = moles of HCl = 0.0923 mol
3. Convert to Volume Using Molarity
If using a 1.0 M NaOH solution: Volume=0.0923 mol1.0 mol/L=0.0923 L=92.3 mL\text{Volume} = \frac{0.0923\ \text{mol}}{1.0\ \text{mol/L}} = 0.0923\ \text{L} = \boxed{92.3\ \text{mL}}
💡 Why This Matters
The purpose of this neutralization is to remove the HCl from Lidocaine HCl, yielding pure, free-base Lidocaine, which is often needed for certain formulations (e.g., transdermal, lipophilic delivery). By adding a stoichiometric amount of NaOH, you ensure the pH is driven just enough to convert the salt to base without excess alkali, which could degrade the product or cause solubility issues.
Using a standard 1.0 M NaOH solution makes this easier to measure accurately. After reaction, extract the free base using a nonpolar solvent (like ether or chloroform) since free Lidocaine is more lipophilic and less soluble in water.
