Question

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calculate the concentration of [Ba 2+] and [ Cl -] following the neutralization reaction.

Answer 1

Answer:

Concentration of the barium ions  = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions  = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride  = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL  = 63.89 mL = 0.06389 L

Concentration of the barium ions =$[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

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:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$