pre lab Experiment 12 THE DETERMINATION OF THE NEUTRALIZING ABILITY OF ANTACIDS
ID: 557328 • Letter: P
Question
pre lab
Experiment 12 THE DETERMINATION OF THE NEUTRALIZING ABILITY OF ANTACIDS The Pre Lab A 1.032 gram sample of an antacid was dissolved in 50 mL of water. Then a 4933 mL portion of 0.488 M HCl was added to the flask. The antacid was completely neutralized and the unneutralized hydrochloric acid remaining in the flask required 29.33 mL of0.510 M NaOH to completely neutralize it. 1. a) The source of all of the acid used in the experiment is the hydrochloric acid, HCI. What are the sources of the two bases that react with this HC1? b) How many moles of the acid, HICI, are used in the experiment? moles of solute volume (in liers) X concentration solute (in molev/iner) c) What must be the total of the number of moles of HCI neutralized by the antacid plus the number of moles of HCl neutralized by the NaOH? d) Calculate the number of moles of the NaOH added during the experiment. moles of solute volume (in liters) X concentration solute (in molesliter) e) How many moles of the total moles of HCI originally present does the added NaOH react with during its neutralization reaction? f) How many moles of the total moles of HCl originally present must the antacid react with during its neutralization reaction? g) How many moles of HCI are neutralized for each gram of antacid tablet used?Explanation / Answer
Solved the complete first problem in detail with sub-parts as per Chegg guidelines, post multiple question to get the remaining answers
Q1)
a) Two bases that react with HCl are NaOH and antacid
b) Moles of acid HCl = Volume of HCl solution in L * Molarity of HCl (M)
=> 49.33/1000 * 0.488 = 0.02407304 moles
c) Number of moles of HCl neutralized by antacid and base is equal to total number of moles of HCl, which is equal to 0.02407304
d) Number of moles of NaOH = Volume of NaOH solution in L * Molarity of NaOH(M)
=> 29.33/1000 * 0.510 = 0.0149583 moles
e) Reaction will be
HCl + NaOH ----- NaCl + H2O
Hence the number of moles of HCl left = number of moles of NaOH = 0.0149583 moles
f) Moles of HCl neutralized by antacid tablet = Total moles of HCl - moles of NaOH
=> 0.02407304 - 0.0149583
=> 0.00911474
g) Moles of HCl neutralized per gram of antacid = 0.00911474/1.032 = 0.0088321124 moles
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