some trick gchem question from destroyer

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datdat

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how many of .12M KOH must be added to 60 ml of 0.2M HF to produce a solution with a pH 3.3(pKa of HF is 3.3)...answer is 5 ml...but don't understand how they got the answer.

any help?
 
hmm not sure, i know that pH = pKa at the half neutralization point, that's whats shown i guess

someone else that knows this please reply, i want to know too
 
something is not right about your question, otherwise people here should have answer it already. You sure it is 5 ml, not 50ml?
 
Last edited:
There's a miscalc in the solution if you look it over again. As it's written:
(.20)(60) = 1.20, which is not correct. It should read (.20)(60) = 12, with V2 = 100 mL (not 10 mL as is listed). The answer then would be 50 mL as opposed to 5 mL. In order for the answer to be 5 mL you would need only 6 mL of .20 M HF. Hope this helps.
 
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