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pH of 1 L 0.1 M HCI solution when 5 g of...

pH of 1 L 0.1 M HCI solution when 5 g of HF is added

A

0.53

B

1

C

0.1

D

0.01

Text Solution

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The correct Answer is:
To find the pH of a 1 L solution of 0.1 M HCl when 5 g of HF is added, we can follow these steps: ### Step 1: Calculate the initial pH of the HCl solution. HCl is a strong acid, which means it completely dissociates in solution. Therefore, the concentration of H⁺ ions from HCl is equal to its molarity. \[ \text{Initial concentration of H⁺ from HCl} = 0.1 \, \text{M} \] Now, we can calculate the pH using the formula: \[ \text{pH} = -\log[H^+] \] Substituting the concentration of H⁺: \[ \text{pH} = -\log(0.1) = 1 \] ### Step 2: Calculate the moles of HF added. To find the effect of adding HF, we first need to calculate the number of moles of HF in 5 g. The molar mass of HF (Hydrofluoric acid) is approximately 20 g/mol. \[ \text{Moles of HF} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \, \text{g}}{20 \, \text{g/mol}} = 0.25 \, \text{mol} \] ### Step 3: Determine the concentration of HF in the solution. Since we are adding HF to 1 L of solution, the concentration of HF will be: \[ \text{Concentration of HF} = \frac{0.25 \, \text{mol}}{1 \, \text{L}} = 0.25 \, \text{M} \] ### Step 4: Analyze the effect of HF on the pH. HF is a weak acid and does not dissociate completely. The dissociation of HF can be represented as: \[ HF \rightleftharpoons H^+ + F^- \] The dissociation constant (Ka) for HF is approximately \(6.8 \times 10^{-4}\). However, since HCl is a strong acid and provides a much higher concentration of H⁺ ions (0.1 M), the contribution of H⁺ ions from HF will be negligible compared to that from HCl. ### Step 5: Conclusion on pH. Since the addition of HF does not significantly increase the concentration of H⁺ ions due to the presence of HCl, the pH will remain very close to that of the original HCl solution. Thus, the final pH will be slightly less than 1, but we can approximate it to: \[ \text{Final pH} \approx 1 \] ### Summary: The pH of the solution after adding 5 g of HF to 1 L of 0.1 M HCl remains approximately 1. ---
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