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Title: How would you find the concentration of a diprotic acid that was titrated with NaOH? Post by: Smoores on Jan 14, 2013 Also use the method to find the concentration of a diprotic acid when 24.55 mL of the acid requires 32.57 mL of 0.1059 M NaOH.
Thank you! Title: How would you find the concentration of a diprotic acid that was titrated with NaOH? Post by: Leona Ariane on Jan 14, 2013 Lancenigo di Villorba (TV), Italy
You are treating two ACID/BASE NEUTRALIZATIONs. TOPICs As you suspected, STOICHIOMETRIC COEFFICIENTs LEAD WORTHY PROPORTION TO REACH NEUTRALIZATION STATUS. For Your Case H2A(aq) + 2 NaOH(aq) ---> Na2A(aq) + 2 H2O(aq) So, once you assured to used only 0.1059 M CONCENTRATED SOLUTION OF STRONG BASE (e.g. NaOH) WHICH OCCURRED IN MEASURE LIKE 32.57 mL AGAINST 24.55 mL OF UNKNOWN ACID. ONCE I WROTE "NET IONIC REACTION" 2 H+(aq) + 2 OH-(aq) ---> 2 H2O(aq) I HIGHLIGHTED THAT NaOH IS MONOVALENT BASE SINCE IT GIVES ONE HYDROXYL ION (e.g. OH- which comes from Arrhenius definition of BASEs). ON THE OTHER HAND, H2A IS BIVALENT ACID SINCE IT GIVES TWO HYDROGEN ION (e.g. H+ which comes from Arrhenius definition of ACIDs). So, YOU PLAYED BY A MONOVALENT BASE AND A BIVALENT ACID. CONCLUSION The mathematical relationship is (n,acid * M,acid) * V,acid = (n,base * M,base) * V,base where n,acid = 2 n,base = 1 V,acid = 24.55 mL V,base = 32.57 mL M,base = 0.1059 M so it has to be M,acid = (n,base * M,base) * V,base / (n,acid * V,acid) = = (1 * 0.1059) * 32.57 / (2 * 24.55) = 0.070 M I hope this helps you. |