If 560mL of 01000 M NaOH solution is needed to just neutrali

If 5.60mL of 0.1000 M NaOH solution is needed to just neutralize excess acid after 20.00 mL of 0.1000 M HCI was added to 1.00 g of an antacid, how many moles of acid can the antacid counteract per gram? mol/g

Solution

Answer:

# of moles = Molarity x Volume of Solution in L.

1) Calculation of Moles of HCl i.e. H⁺ added to an antacid.

Molarity of HCl = 0.1000 M, Volume = 20.00 mL = 0.020 L.

Moles of HCl = Moles of H⁺ = Molarity x volume in L = 0.1000*0.0200 = 0.002 moles.

2) Calculation of Moles of NaOH i.e HO^-. required to neutralize excess H⁺.

Molarity of NaOH = 0.1000 M, Volume of NaOH = 5.60 mL = 0.00560 L.

Moles of NaOH = 0.1 x 0.0056 = 0.00056 moles.

Moles of excess H+ = Moles of HO- required to neutralize excess H+ = 0.00056 moles.

3) Moles of H+ counteracted by An Antacid = Moles of H+ added - Moles of H+ excess

= 0.0020 - 0.00056

= 0.01944 moles.

Moles of Acid counteracted by an antacid = 0.01944 mole/g.

Hence,

0.01944 moles of Acid is counteracted by 1 gram of An Antacid.

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