Question

when the concentration of tartic acid in 10.0mL sample of grape juice if 7.50mL of 0.100 M NaOH solution is needed to neutralize the acid

H2C4H4O6 (aq) + 2NaOH (aq) -------> Na2C4H4O6 (aq) + 2H2O (l)

Answer 1

Moles of NaOH needed for complete neutralization of tartaric acid = C*V = 0.100 mol/L * 7.50 mL * (1L/1000 mL) = 0.00075 mol

In the balanced equation, 1 mol tartaric acid(H₂C₄H₄O₆) is neutralized by 2 mol NaOH.

Hence the mole ratio is: (1 mol tartaric acid / 2 mol NaOH)

=> Moles of rartaric acid in the grape juice sample = 0.00075 mol NaOH * (1 mol tartaric acid / 2 mol NaOH)

= 0.000375 mol tartaric acid

Given volume of grape juice sample, V = 10.0 mL * (1L/1000 mL) = 0.010 L

=> concentration of tartaric acid, C = mol/L = 0.000375 mol / 0.010 L = 0.0375 M (Answer)