Question

Calculate the molarity of each type of ion remaining in solution after 20.0 mL of 6.00 M HCl is mixed with 50.0 mL of 2.00 M barium hydroxide and 30.0 mL of water.

Answer 1

Answer : The molarity of each type of ion remaining in solution are :

Molarity of
`OH^-` ions = 0.8 mole/L

Molarity of
`Cl^-` ions = 1.2 mole/L

Molarity of
`Ba^(2+)` ions = 1 mole/L

Explanation:

The given reaction is,

`HCl + Ba(OH)_(2)\rightarrow BaCl_(2) +H_(2)O`

`Molarity =\frac{\text{moles of solute}}{\text{Volume of solution in liters}}`

`{\text{moles of solute}=\text{molarity}*\text{Volume of solution in liters}}`

`{\text{moles of HCl}=\text{6.00M}*\text{0.02L}}` = 0.12

As 1 mole of HCl dissociates to give 1 mole of
`H^+` and 1 mole of
`Cl^-`

So, 0.12 mole of HCl dissociates to give 0.12 mole of
`H^+` and 0.12
`Cl^-`

`\text{moles of }Ba(OH)_2=2.00M* 0.050L` = 0.1

As 1 mole of
`Ba(OH)_2` dissociates to give 2 moles of
`OH^-` ions,

So, 0.1 moles of
`Ba(OH)_2` dissociates to give 0.1 moles of
`Ba^(2+)` and 0.2 moles of
`OH^(-)`

As 0.12 mole of
`H^+` ions will neutralize 0.12 moles of
`OH^-` ions.

So, remaining
`OH^-` ions in the solution will be 0.2 - 0.12 = 0.08 moles

So, moles of
`OH^-` 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
`OH^-` ions =
`\frac{\text{ moles of }OH^-ions}{\text{ volume of solution in L}}* 1000= (0.08)/(100)* 1000=0.8mole/L`

Molarity of
`Cl^-` ions =
`\frac{\text{ moles of }Cl^-ions}{\text{ volume of solution in L}}* 1000= (0.12)/(100)* 1000=1.2mole/L`

Molarity of
`Ba^(2+)` ions =
`\frac{\text{ moles of }Ba^(2+)ions}{\text{ volume of solution in L}}* 1000= (0.1)/(100)* 1000=1mole/L`