Determine the K_a of a 0.86 M solution of Hydrofluoric acid. HF that is 1.72% io

Question

Determine the K_a of a 0.86 M solution of Hydrofluoric acid. HF that is 1.72% ionized at 25 degree C. Write the equation for the dissociation of an acid and define an expression for K_a Draw an ICE table What other information in the problem above can help you determine the value of x? Use that information to solve for the value of x and substitute back into the ICE table to determine equilibrium concentration of products and reactants Substitute the equilibrium concentration of reactants and products into the K_a expression to determine the value of K_a When you do not know the value of x and have to solve for It. you might be able to make the assumption that x is insignificant However, when you know the value of x in an ICE table, you CANNOT IGNORE SHOULD NOT ignore X. Remember to subtract the x from the Initial concentration In the denominator.

Explanation / Answer

HF <==> H+ +F-

Ka = [F-][H+]/[HF]

Information given that HF is 1.72 % ionized .

(x/0.86) *100 = 1.72

or, x = 0.015

[HF] at equilibrium = 0.86-0.015 = 0.845

[H+] = [F-] = 0.015

Ka = (0.015)*(0.015)/0.845 = 2.66*10^-4

HF H+ F- inital 0.86 0 0 change -x +x +x equilibrium 0.86-x x x

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