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

Concentration of the barium ions =
`[Ba^(2+)] = 0.4654 M`

Concentration of the chloride ions =
`[Cl^(-)]=0.9308 M`

Step-by-step explanation:

`Moles (n)=Molarity(M)* 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*  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:

`(1)/(2)* 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* 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+)]=(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* [Ba^(2+)]=2* 0.4654 M=0.9308 M`