Answer:
Concentration of the barium ions = ![[Ba^{2+}] = 0.4654 M](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%20%3D%200.4654%20M)
Concentration of the chloride ions = ![[Cl^{-}]=0.9308 M](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%5D%3D0.9308%20M%20)
Explanation:

Moles of hydrogen chloride = n
Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L
Molarity of the hydrogen chloride = 0.1355 M


According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.
Then 0.05947 moles of HCl will react with:
barium hydroxide
Moles of barium hydroxide = 0.029735 mol

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

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+}]](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D)
![[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5Cfrac%7B0.029735%20mol%7D%7B0.06389%20L%7D%3D0.4654%20M)

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](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D2%5Ctimes%20%5BBa%5E%7B2%2B%7D%5D%3D2%5Ctimes%200.4654%20M%3D0.9308%20M)