Respuesta :
Answer:
Concentration of the barium ions = [tex][Ba^{2+}] = 0.4654 M[/tex]
Concentration of the chloride ions = [tex][Cl^{-}]=0.9308 M [/tex]
Explanation:
[tex]Moles (n)=Molarity(M)\times Volume (L)[/tex]
Moles of hydrogen chloride = n
Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L
Molarity of the hydrogen chloride = 0.1355 M
[tex]n=0.1355 M\times 0.04389 L=0.005947 mol[/tex]
[tex]Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O[/tex]
According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.
Then 0.05947 moles of HCl will react with:
[tex]\frac{1}{2}\times 0.05947 mol=0.029735 mol[/tex] barium hydroxide
Moles of barium hydroxide = 0.029735 mol
[tex]Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)[/tex]
1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:
[tex]=1\times 0.029735 mol= 0.029735 mol[/tex]
Volume of solution after neutralization reaction :
= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L
Concentration of the barium ions =[tex][Ba^{2+}][/tex]
[tex][Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M[/tex]
[tex]Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)[/tex]
1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.
Then concentration of chloride ions will be:
[tex][Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M[/tex]
