if an antacid tablet is bolied with 50mL of 0.10M HCL, and the resulting solution requires 8.9 mL of 0.095 M NaOH in a back titration, whatr mass of NaHCO3 did the tablet contain ?
Determine the number of moles of HCl added.
50 mL * 1L/ 1000mL * 0.1 mol / 1 L = 0.005 mol HCl
Determine the number of moles of NaOH added.
8.9 mL *1L/1000mL * 0.095 mol / 1 L = 0.0008455 mol NaOH
Determine the number of moles of HCl that reacted with the antacid.
HCl + NaOH ======> NaCl + H2O
0.0008455 mol NaOH * 1mol HCl/1mol NaOH = 0.0008455 mol HCl
0.005 mol - 0.0008455 mol = 0.0041545 mol HCl reacted with antacid
Determine the mass of NaHCO3 in the antacid.
NaHCO3 + HCl =====> H2CO3 + NaCl
0.0041545 mol HCl * 1 mol NaHCO3 / 1 mol HCl * 84.007 g / 1 mol NaHCO3 = 0.349 g NaHCO3 in the antacid.
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