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Answer to Question #57555 in General Chemistry for PRANEET

Question #57555
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
Expert's answer
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

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