Determining the neutralizing power of antacids Given: Weight of antacid tablet:

Question

Determining the neutralizing power of antacids Given: Weight of antacid tablet: 1 g Concentration of HCl: 0.1901 M Volume of HCl added: 100 ml Concentration of NaOH: 0.1032 M Volume of NaOH added: 8.16 ml Volume of the sample (HClI + dissolved tablet) that was titrated using NaOH: 25 ml 1. Calculate the concentration of HClI remaining in the volumetric flask. 2. Calculate the number of moles of HCl that were neutralized by the regular strength TUMS 3. 4. tablet. Calculate the mass of CaCO3 in your TUMS tablet. Determine the standard deviation in your CaCO3 determination.

Explanation / Answer

Balanced equation:
CaCO3 + 2 HCl ===> CaCl2 + CO2 + H2O

Moles of HCl present early = 100 x 0.1901 / 1000 = 0.01901 Moles

Moles of NaOH neutralized by HCl = 8.16 x 0.1032 /1000 = 0.000842112 Moles

Moles of HCl neutralized by TUMS tablet = 0.01901 - 0.000842112 = 0.0181678 Moles

Concentration of HCl =  0.0181678 x 1000 / 108.16 = 0.1679 M

Moles of CaCO3 present = 0.0181678 / 2 = 0.009083944 Moles

Mass of CaCO3 = 0.009083944 x 100.08 = 0.909 gm

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