Mixture by Acid-Base Titration?

Question

Mixture by Acid-Base Titration?
A 1.00-g solid sample containing a mixture of table salt (NaCl) and citric acid (H3C6H5O7, a triprotic acid) is dissolved in 15 mL of water.

Titration of the acid solution requires 13.40 mL of 0.7318 M NaOH solution to reach the endpoint.

Calculate the mass percent H3C6H5O7 in the solid mixture.

_______%

Can Anyone Figure This Out For Me & Explain How They Derived The Answer

Answer 1

From the questions on this site I guess that most students (especially from the US) don't know/understand about gr-eqs and normality so I'll go with moles (and molarity).

H3C6H5O7 + 3 NaOH ->Na3C6H5O7+ 3H2O

from the stoichiometry of the reaction we know that
3 moles of NaOH react with 1 mole H3C6H5O7
y moles NaOH react with x mole H3C6H5O7

Thus 3x=1*y => x= y/3 (1)

but you know that mole=M*V (2)
and also mole= mass/MW (3)

so in equation (1) we substitute x based on equation (2) and y based on equation (3) and we get

mass(H3C6H5O7)/ MW(H3C6H5O7)= M(NaOH)* V(NaOH)/3 =>

mass(H3C6H5O7)= MW(H3C6H5O7)* M(NaOH)* V(NaOH)/3=
= 192*0.7318* 13.40* (10^-3)/3= 0.6276 gr

So the % is mass(H3C6H5O7)/mass (sample)= 0.6276/1= 62.76% H3C6H5O7

Answer 2

overlook with regard to the NaCl because it doesnt play any section in the reaction. this is going to take 3 moles of NaOh to react with a million mole of citric acid thoroughly. artwork out ho wmany moles of NaOH you have then divide it by way of 3 to get moles of citric.