Question

Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 solution as Mn(OH)2. The equation for the reaction is: MnSO4(aq) + Ba(OH)2(aq) Mn(OH)2(s) + BaSO4(aq)

Answer 1

Answer: The volume of barium hydroxide is 183 mL.

Step-by-step explanation:

To calculate the moles of cadmium nitrate, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}` .....(1)

Molarity of
`MnSO_4` = 0.796 M

Volume of
`MnSO_4` = 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

`0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol`

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

`MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)`

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with =
`(1)/(1)* 0.13=0.13mol` of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

`0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L`

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So,
`0.183L=0.183* 1000=183mL`

Hence, the volume of barium hydroxide is 183 mL.