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.
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?
2 answers
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?
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?
1 answer