Homework Help: chemistry

Posted by Anonymous on Friday, November 4, 2011 at 1:19am.

a student dissolves 0.625 g of pure benzoic acid in distilled water and titrated the resulting solution to the equivalence point using 40.8 mL of the standardized NaOH solution from Part a, assuming that benzoic acid has only one ionizable hydrogen, answer the following: writing the unknown formula of benzoic acid as HA, where H is the ionizable hydrogen atom, write the balanced molecular equation that occurs in this titration.. Part A: the student begins by standardizing a solution of NaOH. he uses potassium hydrogen phthalate, KHC8H4O4 (MM=204.22 g/mol), a monoprotic acid which reacts with NaOH as shown in the following balanced equation: KHP + NaOH = H2O =KNaP

chemistry - DrBob222, Friday, November 4, 2011 at 8:22pm
You don't have questions here. Mostly just statements. Here is the equation.
HA + NaOH ==> H2O + NaA

I assume the benzoic acid was titrated with the NaOH and the question is to determine the molar mass of benzoic acid.
LNaOH x M NaOH (from part a) = moles NaOH.
moles HA = moles NaOH
Then moles HA = grams HA/molar mass HA.
You have moles HA and grams HA, solve for molar mass.

From part a.
Here is what you do with the data you have to determine M NaOH. (No data is given in the post.)
moles KHP = grams/molar mass = grams/204.22.
moles NaOH = moles KHP since the rxn is 1:1.
M NaOH = moles NaOH/L NaOH.
You have moles NaOH and you have the titrated volume between KHP and NaOH which is L. Solve for M NaOH and use that in the benzoic acid titration you describe first.

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