25.7 mL of 0.13 M HCl is added to 0.67 g or commercial antacid. Some, not all, o

Question

25.7 mL of 0.13 M HCl is added to 0.67 g or commercial antacid. Some, not all, of the added acid is neutralized by the antacid. Phenolphthalein is added, and the solution remains colorless because of the excess acid that has not been neutralized. The solution titrated with 8.9 mL of 0.11 M NaOH to a faint pink endpoint. If an entire tablet of this antacid has a mass of 1.46 g, how many moles of HCl will be neutralized by one tablet of this antacid? can you please show your work for this problem. thank you

Explanation / Answer

First let figure out how many mols of acid we added .0257 L * 0.13 mol/L =0.003341 moles of acid Now some of that acid reacts with the commercial antacid, but some is left over. To find out how much was left over, we calculate the moles of NaOH like this: 8.9 mL * 0.11 mol/L = 0.000979 moles of NaOH Now we can tell how much of the acid reacted with the antacid by taking the difference of these numbers: 0.003341 - 0.000979 = 0.002362 moles of acid So we know that 0.67 g of antacid neutralized 0.002362 mols of acid. And we know that the mass of the entire tablet is 1.46 g. So we set up a ratio and cross multiply: 0.002326 mol / 0.67 g = x mol / 1.46 g x = 0.002326 mol * 1.46 g / 0.67 g = 0.005069 mols of acid coul be neutralized by the whole tablet.

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