Question

# A student dissolves 0.326 g of a powdered antacid in 32.36 mL of 0.1034 M HCl....

A student dissolves 0.326 g of a powdered antacid in 32.36 mL of 0.1034 M HCl. The student boils the mixture and then allows it to cool. Lastly, the student adds phenolphthalein indicator to the mixture.

3) Calculate the difference between total moles of H+ added and total moles of OH- added.

It is Back titration , Here excess NaOH is added to sample intially in it some amount of NaOH consumed for neutralize sample , remaining unreacted is calculated by titration with standard HCl.

So from difference between Volume of NaOH and unreacted NaOH gives amount of NaOH consumed for reacting with antacid.

moles of NaOH in the titration = 0 .1506 x 0.01172 = 0.001765 moles

Because NaOH + HCl = NaCl + H2O the moles of excess HCl =0 .001765 moles

moles of HCl originally added =0 .1034 x 0.03236 = 0.003346

so the moles of HCl that reacted with at antacid = the difference
= 0.001580 moles of HCl

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