Question

A 20.00 mL wine sample is titrated with a 1.04×10^-1M solution of sodium hydroxide. The equivalence point is determined to be 25.926 mL. What is the acidity of the wine expressed as (g tartaric acid / L; MW tartaric acid = 150.085 g/mole)?

Answer 1

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OH - x1 le mole = mass o Molar Mass of Tantulle aideHy C4 H406 (Hot Sodium hydroxide (Nroot? The balance equation of the reaction by these two componente H₂ CQH4O6 cag) t 2 NaOH cag) - NaC H4 6 car) + 2 bore) mol mol flere, NaOH 1604x16M = 0.104M = 0.104 mollt Volume = 25.926 mL = 0.0259264 Molasite = mass of solute (9) molar mass of solute glad) volume of solution (L) Molarity = no. of moles of solulé (mol) volume of solution (L) (** #M = ņ n= MXV = 6.104 mol/L) (0.025926L) = 0.00969 mol from equation, 1 mal Hoof react with a mol Naot – mol that of reacts with 0.00269 mol Naon mol x 0.00269 mol = 0.001348 mol. amel calculate the mass of Halattoo mass = no of mol x molar mass = 0.001348 molx 150.085g/mol = 0.20239 of tartaric acid. This tartarie and in present in 20ml or 0.020L of sample. concentration = 0.20234 = loalt g tastasie ace/L 0.020L