Excess 0104275 M HCl 5000 mL was added to 08204 g of an unkn

Excess 0.104275 M HCl (50.00 mL) was added to 0.8204 g of an unknown sample of sodium bicarbonate (MM: 105.988 g/mol). 14.50 mL of 0.14265 MNaOH was used to complete the titration.

Calculate the percentage of sodium bicarbonate in the sample.

Solution

Ans. Step 1: Neutralization of excess HCl with NaOH

Balanced reaction:    HCl(aq) + NaOH(aq) ----> NaCl(aq) + H₂O(l)

According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol HCl.

# Moles of NaOH consumed = Molarity x Volume of solution in liters

                                                = 0.14265 M x 0.01450 L

                                                = 0.002120625 mol

So, moles of excess HCl = Moles of NaOH consumed = 0.002120625 mol

# Step 2: Total initial moles of HCl taken = 0.104275 M x 0.050 L = 0.00521375 mol

Mol of HCl consumed to neutralize NaHCO₃ = Total HCl moles – Moles of excess HCl

                                                = 0.00521375 mol - 0.002120625 mol

                                                = 0.003093125 mol

# Step 3: Neutralization of NaHCO₃ with HCl

Balanced reaction:    HCl(aq) + NaHCO₃(aq) ----> NaCl(aq) + H₂O(l) + CO₂(g)

According to the stoichiometry of balanced reaction, 1 mol NaHCO₃ neutralizes 1 mol HCl.

So,

Moles of NaHCO₃ in sample = Moles of HCl consumed by NaHCO₃ sample

                                                = 0.003093125 mol

# Now, Mass of NaHCO₃ in sample = Moles x Molar mass

                                                = 0.003093125 mol x (84.0 g/ mol)

                                                = 0.2598225 g

# % NaHCO₃ = (Mass of NaHO₃ / Mass of sample) x 100

                                    = (0.2598225 g / 0.8204 g) x 100

                                    = 31.67 %