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adam160596 Group Title

What would the pH of 0.56M HF be? How would I go about solving this

  • 3 months ago
  • 3 months ago

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  1. shoba Group Title
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    is there any other information given..? pka value?

    • 3 months ago
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    HF <=>H+ + F- initial no of mols HF = 0.56 H+ = 0 F- = 0 at equilibrium HF = 0.56 - x H+ = x F- = x ka = [H+] * [F-] / [HF] pka = -log ka from that you can find ka and ka = x * x / (0.56 - x) from this x can be calculated. x is the H+ mols and since we got to the 1000 cm3 (we simply got the concentration as no of mols) x = [H+] then ph can be calculated. ---------------------------------------... ka = 10^ -3.14 ka = 7.244 * 10 ^-4 7.244 * 10 ^-4 = x^2/ [0.56-x] x is very small so 0.56 - x can be take as 0.56 7.244 * 10 ^-4 = x^2/ 0.56 x^2 = 0.0004032 x = 0.0200798 = [H+] pH = -log [H+] pH = 1.697 *assumed the pka value 3.14

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