Question

During the quick titration of 10 ml of lemon juice with 0.1053 mol/L NaOH, a student dispensed 82mL of sodium hydroxide (titrant) into the lemon juice (anylate) before the end point was reached. Estimate the anylate concentration assuming the acid in the lemon juice is monoprotic

Answer 1

Acid is monoprotic, so n(acid) = n(NaOH)
n(NaOH) = c*V = 0.1053 * 0.082 = 8.6*10-3 (mol) = n(acid)
c(acid) =n/V = 8.6*10-3 / 1*10-2 = 0.86 (mol/L)

Answer
c(acid) = 0.86 mol/L