Asked by Gordon

A 5.0ml sample of liquid requires 4.5 ml of a 0.093N NaOH solution. What is the titratable acidity of the liquid expressed as g/L Tartaric acid?
Answered by DrBob222
mL x N x mew = grams tartaric acid.
mew = milliequivalent weight of tartaric acid = molar mass/2 = ?
grams from above is in 5.00 mL. Convert that to g/L.
Answered by Gordon
mew = HOOC-CHOH-CHOH-COOH/2
150 gr/mol /2 = 75gr/mol

4.50ml x .093N x 75 gr/mol = 31.4 grams of tartaric acid in the 5ml sample.

31.4gr /5.0 ml = x gr/1 liter

is this correct?



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