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adam160596
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What would the pH of 0.56M HF be? How would I go about solving this
 3 months ago
 3 months ago
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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shoba Group TitleBest ResponseYou've already chosen the best response.1
is there any other information given..? pka value?
 3 months ago

shoba Group TitleBest ResponseYou've already chosen the best response.1
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/ [0.56x] 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
 3 months ago
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