ASA: Exp6 - Acid Neutralizing Power of Commercial Antacids 3/29 & 4/1/16 Using t
ID: 1035473 • Letter: A
Question
ASA: Exp6 - Acid Neutralizing Power of Commercial Antacids 3/29 & 4/1/16 Using the data given below, complete the following calculations based on the description given in the lab procedure. There are a few things that you need to keep in mind about this titration. First, this is a backward titration, which means that you completely react the chemical that you are most interested in with an excess of acid (sometimes a base), then you titrate the acid (or base) that's left over to determine how much ofyour chemical of interest was present. You titrating that chemical poses problems like the CaCOa that we're titrating today (i.e. It forms a buffer which messes up the titration curve making the equivalence point & endpoint really hard to determine.) Or you can do a backward titration ifyou mess up a normal titration by passing the equivalence or endpoint. FINALLY at the equivalence point (which should be nearly identical to the do this because chose the correct indicator), the # moles of the titrant Nao rus: # mole e chemical being titrated Given: HCl]stock 0.1 M: Caco,(aq) + 2 HCI(aq) CaC12(aq) + H20() + CO2(g) Calculation: A student adds 0.355 g of powdered tablet (containing some CaC03) to 50 mL or HClstock. [NaOH]stock-0.1 M: 1 tablet-2.640 g Tablet(total) mmol HCladded (Total!y After heating this solution to drive off any CO2g) to prevent buffer formation then cooling it, the excess HCl in the acidified solution was titrated with 23.12 mL of Na0Hstock. mmol Na0Hadded mmol HClttrated (excess) mmol HClreacted (with tablet) mmol CaCO3 (in the reacted tablet) mg CaCO3 (in the reacted tablet) According to the chemical equation above, Considering: mg CaC02(reacted) g Tablet(reacted) mg GaC03 Tablet Given that the reported value is 1050mgCacOs Tablet' % errorExplanation / Answer
The no. of mmol of HCl added = 50 mL * 0.1 mmol/mL = 5 mmol
The no. of mmol of NaOH added = 23.12 mL * 0.1 mmol/mL = 2.312 mmol
The no. of mmol of HCl titrated (excess) = 2.312 mmol
The no. of mmol of HCl reacted = 5 - 2.312 = 2.688 mmol
The no. of mmol of CaCO3 in the reacted tablet = 1/2 * 2.688 = 1.344 mmol
The mass of CaCO3 in the reacted tablet = 1.34 mmol * 100 g/mol = 134 mg
mg CaCO3/tablet = (134 mg/0.355 g)*2.64 g = 996.5 mg CaCO3/tablet
% error = 100 - {(996.5/1050)*100} = 5.1%