Asked by sam
to drbob222
ok, the pka (hclo) = 7.54
the ph = 7.35
this problem was under "preparing a buffer" section
ok, the pka (hclo) = 7.54
the ph = 7.35
this problem was under "preparing a buffer" section
Answers
Answered by
DrBob222
7.35 = 7.54 + log(base)/(acid)
(base)/(acid) = 0.646
I finally found the problem again. It says you have 500 mL of 0.1M HClO which is 50 mmoles and wants to know how much NaClO would be added.
NaClO/HClO = 0.646
NaClO = 0.646*50 mmols = ??
I would then divide by 1000 to convert to mols.
moles NaClO = grams/molar mass
You know moles NaClO and you know molar mass, solve for grams.
I like to check these things by putting moles NaClO and moles HClO back into the HH equation and see if it gives the correct pH.
(base)/(acid) = 0.646
I finally found the problem again. It says you have 500 mL of 0.1M HClO which is 50 mmoles and wants to know how much NaClO would be added.
NaClO/HClO = 0.646
NaClO = 0.646*50 mmols = ??
I would then divide by 1000 to convert to mols.
moles NaClO = grams/molar mass
You know moles NaClO and you know molar mass, solve for grams.
I like to check these things by putting moles NaClO and moles HClO back into the HH equation and see if it gives the correct pH.
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