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
1
Expert's answer
2016-01-29T09:46:55-0500
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

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS