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dissociation of HF
I am unsure as to how to obtain the correct answer
Calculate [H+] and the % dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2x10^-4) and 1.0 M NaF? do I make an ICE table with: HF + NaOH <-> H2O + NaF 1 1 -x +x It just doesn't make sense to me. I think I have the wrong initial equation. I know how to do % dissociation. I just can't get started. Any help please2 |
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NaF is a strong electrolyte. So assume that its dissocation goes to completion and you'll have [F-]= 1 M. The equlibrium equation is HF <-> H+ + F- I 1.0 0 1.0 C -x x +x E 1-x x 1+x Then use definition of K. K = (1+x)(x)/(1-x). Solve for x. That's your answer ! |
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