The volume of 0.1 M HCl required in mL to neutralise 20 ~mL of a solution containing 0.106 ~g of Na ₂ CO ₃ is

Step-by-step solution

Given, Molarity ( HCl ) [M₁ ]=0.1 M Volume of Na ₂ CO ₃ used =20 ~mL Mass of Na ₂ CO ₃ used =0.106 ~g thus, molarity (M₂ ) of Na ₂ CO ₃= 0.106 106 1000 20 (M₂ ) Na ₂ CO ₃=0.05 M Equation for neutralisation is M₁ V₁ Z₁=M₂ V₂ Z₂ Where Z₁ and Z₂ are number of moles of H ⁺ and ions, given by HCl and Na ₂ CO ₃ in water aligned & Z₁=1( for HCl ) ; Z₂=2 ( for Na ₂ CO ₃ ) & V₁= M₂ V₂ Z₂ M₁ Z₁ = 0.05 20 2 0.1 1 =20 ~mL aligned Hence, volume of HCl required =20 ~mL