Final answer:
To calculate the molar concentration of HF from a mixture containing NH4F and NH4HF2, stoichiometry and the dissociation reactions are used. Given the molarities, NH4F does not contribute to HF concentration, while NH4HF2 does directly, giving a total HF molarity of 1.64 M in the solution.
Step-by-step explanation:
In order to calculate the weight percent expressed in HF or the molar concentration of HF from a mixture of NH4F and NH4HF2 in water, one would first need to understand the dissociation reactions of the compounds in water.
Ammonium fluoride (NH4F) dissociates into NH4+ and F- ions, while ammonium bifluoride (NH4HF2) dissociates into NH4+ and HF2- ions, which can further release HF.
Using the molarities given for NH4F and NH4HF2, you could apply stoichiometry to the dissociation reactions. Since NH4F releases one mole of fluoride ions per mole of NH4F, the molarity of fluoride ions in solution would be equal to the molarity of NH4F.
For NH4HF2, one mole of NH4HF2 releases one mole of HF upon dissociation, so the molarity of HF from NH4HF2 will be the same as the molarity of NH4HF2.
Calculating HF Concentration
To find the total molarity of HF in the solution, add the molarity contributions from both the NH4F and NH4HF2 based on their dissolution reactions and the given molarities:
- From NH4F: 0 M (since NH4F does not produce HF directly)
- From NH4HF2: 1.64 M (since NH4HF2 dissociates directly to give HF)
Total HF concentration would thus be the sum of contributions which equals 1.64 M. This calculation assumes 100% dissociation for simplicity. To further convert to weight percent, you would need the total mass of the solution and the specific mass contribution from HF.