Neutralization of Antacid A. Standardization ofNaoH Molarity of HCI M Initial re
ID: 511534 • Letter: N
Question
Neutralization of Antacid A. Standardization ofNaoH Molarity of HCI M Initial reading of HCI Final reading ofHCI Volume of HCI Initial reading of NaOH Final reading of NaOH Volume of NaOH Molarity of NaoH (eq. 20 Mean molarity of NaoH B. Back-Titration of Antacid Mass of antacid powder Initial reading of HCI Final reading ofHCI Volume of HCI Initial reading ofNaoH Final reading of NaOH Volume of NaOH H" neutralized (eq. 1) Caco (in Tums) neutralized (eq. 3) Molar mass of Caco, Mass of CaC0, CaCO Trial 3 Trial 2 Trial 1 U.21 mL 20.18 mL, 04 Trial 2 Trial 1 50 LIU 40 mL. mL 75.0 25.3a l 2, 34 .02. mmol mmol mmol mol grmol 9%Explanation / Answer
Volume of NaOH :
in trial 1 : V = 13.34 mL
in trial2 : V = 16.02 mL
Moles of NaOH used :
in trial 1 =0.094 M * 0.01334 L = 1.25*10^-3 moles
in trial 2 = 0.094 M * 0.01602 L = 1.51*10^-3 moles
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H+ neutralized in
Trial 1 : 1.25 mmoles
Trial 2 : 1.51 mmoles
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mmoles of CaCO3 in
Trial1 :1.25 mmol
Trial2 : 1.51 mmol
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Molar mass of CaCO3 = 100gm/mol
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Mass of CaCO3 :
Trial 1 = 1.25 *10^-3 mol *100 g/mol = 0.125 gm
Trial 2 = 1.51 *10^-3 *100g/mol = 0.151 gm
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% mass of CaCO3
trial 1 : (0.125 g/0.503) *100% = 24.85 %
trial2 : (0.151/0.705) *100 % = 21.42 %