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Fynjy0 [20]
3 years ago
6

25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti

on has been added, the pH in the titration flask is 3.00. What was the concentration of the original hydrofluoric acid solution. (Ka(HF) = 7.1 × 10–4)
Chemistry
1 answer:
lesantik [10]3 years ago
6 0

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3.   This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

                HF  ⇄    H⁺    +      F⁻

C(eq)       [HF]     10⁻³M      0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

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it takes 463 kJ/mol to break an oxygen-hydrogen single bond. calculate the maximum wavelength of light for which an oxygen-hydro
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Answer:

  • <u>2.59 × 10⁻⁷ m  = 259 nm</u>

Explanation:

You need to calculate the wavelength of a photon with an energy equal to 463 kJ/mol, which is the energy to break an oxygen-hydrogen atom.

The energy of a photon and its wavelength are related by the Planck - Einstein equation:

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Where:

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And:

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Where:

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  • λ = wavelength of the photon

Thus, you can derive:

  • E = h c / λ

Solve for λ:

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Substitute the values and use the energy of one bond:

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The wavelength of light is usually shown in nanometers:

  • 2.59 × 10⁻⁷ m × 10⁹ nm / m = 259 nm ← answer
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