Question

8.4 MG nahco3 powder was slowly added to 10 ml of 0.03 m HCL in a 50 ml breaker steering after effervescent seized

Answer 1

We have given,

weight of NaHCO₃ = 8.4 mg

molecular weight of NaHCO₃ = 84 g/mol

Number of moles of NaHCO₃ \(=\frac{0.0084\,g}{84\,g/mol}\)

= 0.0001 mol

Concentration of HCl = 0.03 M

Volume of HCl = 10 mL

∴ moles of HCl = 0.03 M x 0.01 L

= 0.0003 mol

NaHCO₃ + HCl → NaCl + H₂O + CO₂ ↑

∵ 1 mole of NaHCO₃ reacts with 1 mol. of HCl and give 1 mol NaCl, 1 mol H₂O and 1 mol CO₂ gas.

∴ 0.0001 mol, NaHCO₃ reacts with 0.0001 mol HCl

So, left HCl = 0.0002 mol

∴ Concentration of HCl = 0.0002/10 x 1000

= 0.02 M HCl