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Hey DrBoB222 I was wondering if you could help me with that problem just a little bit ago again?
2 answers
Of course, I'd be happy to help! Please go ahead and ask your question, and I'll do my best to assist you.
Answered below.
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DrBob222,
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to drbob222
ok, the pka (hclo) = 7.54 the ph = 7.35 this problem was under "preparing a buffer" section
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7.35 = 7.54 + log(base)/(acid) (base)/(acid) = 0.646 I finally found the problem again. It says you
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solve Ka for HX => H^+ + X^- given 0.1M LiX at pH=8.9
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yes. It is 0.1 initially, it is 0.1-x at equilibrium in the ICE chart.
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