Respuesta :
Answer : The molarity of each type of ion remaining in solution are :
Molarity of [tex]OH^-[/tex] ions = 0.8 mole/L
Molarity of [tex]Cl^-[/tex] ions = 1.2 mole/L
Molarity of [tex]Ba^{2+}[/tex] ions = 1 mole/L
Explanation:
The given reaction is,
[tex]HCl + Ba(OH)_{2}\rightarrow BaCl_{2} +H_{2}O[/tex]
[tex]Molarity =\frac{\text{moles of solute}}{\text{Volume of solution in liters}}[/tex]
[tex]{\text{moles of solute}=\text{molarity}\times\text{Volume of solution in liters}}[/tex]
[tex]{\text{moles of HCl}=\text{6.00M}\times\text{0.02L}}[/tex] = 0.12
As 1 mole of HCl dissociates to give 1 mole of [tex]H^+[/tex] and 1 mole of [tex]Cl^-[/tex]
So, 0.12 mole of HCl dissociates to give 0.12 mole of [tex]H^+[/tex] and 0.12 [tex]Cl^-[/tex]
[tex]\text{moles of }Ba(OH)_2=2.00M\times 0.050L[/tex] = 0.1
As 1 mole of [tex]Ba(OH)_2[/tex] dissociates to give 2 moles of [tex]OH^-[/tex] ions,
So, 0.1 moles of [tex]Ba(OH)_2[/tex] dissociates to give 0.1 moles of [tex]Ba^{2+}[/tex] and 0.2 moles of [tex]OH^{-}[/tex]
As 0.12 mole of [tex]H^+[/tex] ions will neutralize 0.12 moles of [tex]OH^-[/tex] ions.
So, remaining [tex]OH^-[/tex] ions in the solution will be 0.2 - 0.12 = 0.08 moles
So, moles of [tex]OH^-[/tex] ions after the reaction = 0.08 moles
Now we have to calculate the molarity of each type of ions remaining in solution.
Total volume = 20 + 30 + 50 = 100 ml
Molarity of [tex]OH^-[/tex] ions = [tex]\frac{\text{ moles of }OH^-ions}{\text{ volume of solution in L}}\times 1000= \frac{0.08}{100}\times 1000=0.8mole/L[/tex]
Molarity of [tex]Cl^-[/tex] ions = [tex]\frac{\text{ moles of }Cl^-ions}{\text{ volume of solution in L}}\times 1000= \frac{0.12}{100}\times 1000=1.2mole/L[/tex]
Molarity of [tex]Ba^{2+}[/tex] ions = [tex]\frac{\text{ moles of }Ba^{2+}ions}{\text{ volume of solution in L}}\times 1000= \frac{0.1}{100}\times 1000=1mole/L[/tex]
