Asked by rose
if a 240mg of an acid H2A required 50.0ml of 0.10M NaOH for complete neutralization, what is yhe equivalent weight of the acid? what is its molecular weight?
Answers
Answered by
DrBob222
0.1M NaOH = 0.1N NaOH
H2A + 2NaOH ==> Na2A + 2H2O
When the problem says "complete" neutralization, that mans BOTH H atoms have been replaced.
mL x N x m.e.w = grams.
50.00 x 0.1N x m.e.w. = 0.240g
solve for m.e.w = milliequivalent weight.
Multply by 1000 to convert to equivalent weight.
Then equivalent weight = molecular weight/replaceable Hs.
You know equivalent weight and replaceable H atoms (2), solve for molecular weight.
H2A + 2NaOH ==> Na2A + 2H2O
When the problem says "complete" neutralization, that mans BOTH H atoms have been replaced.
mL x N x m.e.w = grams.
50.00 x 0.1N x m.e.w. = 0.240g
solve for m.e.w = milliequivalent weight.
Multply by 1000 to convert to equivalent weight.
Then equivalent weight = molecular weight/replaceable Hs.
You know equivalent weight and replaceable H atoms (2), solve for molecular weight.
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