Question

On a separate piece of paper (not in your lab notebook), show what volume of concentrated HCl (12 M) is needed to fully neutralize 5 mL of a 1 M solution of sodium bicarbonate (show your setup and your work). Also write the balanced chemical equation for this neutralization reaction.

Answer 1

The reaction of fully neutralization of sodium bicarbonate by HCl is as follows-

HCl(aq) + NaHCO₃(aq) = NaCl(aq) + H₂O(l) + CO₂(g)

So one mole of HCl will completely neutralize one mole of NaHCO₃.

According to the conservation of activity,

V₁ × S₁ = V₂ × S₂

[where V₁=volume of HCl = ? and S₁ = concentration of HCl = 12M; V₂ = volume of NaHCO₃ = 5mL; S₂ = concentration of NaHCO₃ = 1M]

So, V₁ = (V₂ × S₂)/S₁ = (5mL × 1M)/12M = 0.417mL

So volume of concentrated HCl (12M) required for fully neutralization of 5mL of 1M NaHCO₃ is 0.417mL.