Answer:
20mL.
Explanation:
The volume of 0.125 M HF needed to react with 50 mL of 0.05 M NH₃ solution is determined. The reaction involves neutralization:
[tex] \text{HF} + \text{NH}_3 \rightarrow \text{NH}_4^+ + \text{F}^- [/tex]
This is understood from the stoichiometry of the balanced equation, where the molar ratio of HF to NH₃ is 1:3. In other words, 1 mole of HF reacts with 1 mole of NH₃.
So, let's first calculate the moles of NH₃ that are present in a solution:
[tex] \text{moles of NH}_3 = \text{Molarity} [/tex]
[tex] \text{MoL of NH}_3 = 0.05 \, M \times 0.050 [/tex]
[tex] \text{Moles of NH}_3 = 0.0025 [/tex]
Since the reaction ratio is 1:1, then required moles of HF also will be 0.0025 moles. Now we have to find the volume of 0.125 M HF solution containing 0.0025 moles of HF:
[tex] \text{Volume of HF} = \dfrac{\text{Moles of HF}}{\text{Molarity of HF}} [/tex]
[tex] \text{Volume of HF} = \dfrac{0.0025 \, \text{moles}}{0.125 \, \text{M}} [/tex]
[tex] \text{Volume of HF} = 0.02 \, \text{L} [/tex]
Change the volume from liters to milliliters:
[tex] \text{Vol. of HF} = 0.02 [/tex]
[tex] \text{Volume of HF} = 20 \, \text{mL} [/tex]
Therefore 20 mL of 0.125 M HF is titrated against 50 mL of 0.05 M NH₃ solution.
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