Titration Techniques using NaOH (0.110M), HCl (0.700M), and Alka-Seltzer tablet

Question

Titration Techniques using NaOH (0.110M), HCl (0.700M), and Alka-Seltzer tablet (total mass of 1 tablet Alka-Seltzer = 3.2646g)

Alka Seltzer-- Active Ingredients: Sodium bicarbonate (1916 mg), Aspirin (325 mg), Anhydrous Citric Acid (1000mg)

(1) Calculate the moles of acid that were neutralized by the piece of antacid tablet.

(2) Use the moles of acid neutralized by the piece of tablet to calculate the moles of acid that could be neutralized by the entire antacid tablet.

For #1, I Calculated:

Trial 1- 0.0061730746 mol HCl

Trial 2- 0.0047803833 mol HCl

Trial 3- 0.0070717772 mol HCl

I can't figure out where to begin for Question #2. I need help setting up for #2 and what the answers would be so that I can compare once I try to calculate it on my own.

Trial 1 Trial 2 Trial 3 Mass of piece of tablet 0.5186 g 0.4016 g 0.5941 g V(total) NaOH 12.19 mL 9.18 mL 11.05 mL V(total) HCl 32.83 mL 25.98 mL 35.3 mL

Explanation / Answer

Active Ingredient that neutralizes acid in stomach of human is Sodium bicarbonate (1916 mg).(Antacid)

total mass of antacid in the tablet=1916 mg* (1g/1000mg)=1.916g

mol of sodium bicarbonate in the entitire talet=mass/molar mass of sod bicarbonate=1.916g/(84.007g/mol)=0.02281 mol

(1) trial 1)

mass of tablet piece taken=0.5186 g

Excess HCl is added to tablet mass,to neutralize the antacid (Sodium bicarbonate) completely.

mol of HCl added=0.03283 L *(0.7mol/L)=0.02298 mol

mol of NaOH required for back-titration of excess HCl=0.01219 L* 0.110M=0.001341 mol

mol of HCl neutralized in back -titration=0.001341mol

Thus, mol of HCl required for neutralization of sodium bicarbonate=0.02298 mol-0.001341mol=0.021639 mol

NaHCO3 + HCl -> NaCl + H2O + CO2 (neutralization rxn)

mol NaHCO3/mol HCl)=1/1

mol of sodium bicarbonate actually present in the tablet piece=(mol of HCl)=(0.021639 mol)

Thus mol of HCl(neutralized by the tablet piece=mol of sod bicarbonate=0.021639 mol

(2)

mol of acid neutralized by 0.5186 g tablet piece=0.021639 mol

So, mol of acid neutralized by the entire tablet=(0.021639 mol/0.5186g tablet piece)*3.2646g =0.1362 mol

trial #2)

mass of tablet piece taken=0.4016 g

Excess HCl is added to tablet mass,to neutralize the antacid (Sodium bicarbonate) completely.

mol of HCl added=0.02598 L *(0.7mol/L)=0.01819 mol

mol of NaOH required for back-titration of excess HCl=0.00918 L* 0.110M=0.00101 mol

mol of HCl neutralized in back -titration=0.00101mol

Thus, mol of HCl required for neutralization of sodium bicarbonate=0.01819 mol mol-0.00101mol=0.01718 mol

mol NaHCO3/mol HCl)=1/2

mol of sodium bicarbonate actually present in the tablet piece=mol of HCl)=0.01718 mol

Thus mass of acid neutralized=mol of bicarbonate=0.01718mol

(2)

mol of acid neutralized by 0.4016 g tablet piece=0.01718mol

So, mol of acid neutralized by the entire tablet=(0.01718molmol/0.4016g tablet piece)*3.2646g =0.1396 mol

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