Titration Question
I need help with working out the Mr of Succinic Acid and the Concentration of the Acid!
Equation: C4H6O4 + 2NaOH -------> Na2C4H4O4 + 2H2O
Volume of Succinic Acid = 25cm3
Average Titre of NaOH = 25.60cm3
Concentration of NaOH= 0.1moldm-3
Mass of Acid= 1.51g
Equation: C4H6O4 + 2NaOH -------> Na2C4H4O4 + 2H2O
Volume of Succinic Acid = 25cm3
Average Titre of NaOH = 25.60cm3
Concentration of NaOH= 0.1moldm-3
Mass of Acid= 1.51g
(edited 9 years ago)
Original post by S242
mol of NaOH=(25.60/1000)*1=0.0256mol
2NaOH:1C4H6O4
so,no.of mol of succinic acid=(1/2)*0.0256=0.0128mol
no. of mol of acid=mass of acid/Mr of acid
Mr =mass/no.of mol.
Mr=1.51/0.0128
Mr=118g/mol
concentration of acid=(no.of mol *1000)/volume
concentration=(0.0128*1000)/25=0.512mol/dm3
2NaOH:1C4H6O4
so,no.of mol of succinic acid=(1/2)*0.0256=0.0128mol
no. of mol of acid=mass of acid/Mr of acid
Mr =mass/no.of mol.
Mr=1.51/0.0128
Mr=118g/mol
concentration of acid=(no.of mol *1000)/volume
concentration=(0.0128*1000)/25=0.512mol/dm3
Thanks very much for verifying my answer!
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