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Data and Calculation (show all calculations): Name of the antacid brand Active i

ID: 504875 • Letter: D

Question

Data and Calculation (show all calculations): Name of the antacid brand Active ingredient from label Amount of HCI added to the antacid, 0.00200 L times 0.100 M mole NaOH titration against excess HCI Mole NaOH = M times V (in L) = (0.100) L NaOH used) HCI used in neutralizing the antacid (base): 0.200 - NaOH mole used in titration HCI___ mole ___ mole ___ mole Mass of antacid weighed ___ g ___g ___g Neutralizing power of antacid, mole of HCI (stomach acid)/g antacid ___ ___ ___ Average neutralizing power, mole HCI/g antacid ___

Explanation / Answer

CaCO3 + 2HCl ---> CaCl2 + H2CO3

NaOH + HCl ---> NaCl + H2O

Total HCl added = 0.1 M x 2 ml = 0.2 mmol

moles NaOH = grams NaOH/40 g/mol

                                Trial 1                Trial 2              Trial 3

NaOH (mmol)            0.5                     0.5                   0.5

Since moles of NaOH is greater than the total moles of HCl added, some discreapancy exists in the volume of HCl and/or mass of NaOH used.

moles of HCl used for neutralizing = total moles of HCl added - moles of NaOH used

From this data,

neutralizing power of antacid, which is = moles of HCl/grams of antacid

is calculated

From three trial runs, average value can be calculated easily.