Question

A student finds that 24.52 mL of 0.7170 M silver nitrate is needed to precipitate all of the bromide ion in a 50.00-mL sample of an unknown. What is the molarity of the bromide ion in the student's unknown?

Answer 1

Answer: The molarity of bromide ions is 0.348 M.

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 silver nitrate = 0.7170 M

Volume of silver nitrate = 24.52 mL = 0.02452 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

`0.7170mol/L=\frac{\text{Moles of silver nitrate}}{0.02425L}\\\\\text{Moles of silver nitrate}=0.0174mol`

The chemical equation for the reaction of silver nitrate and bromide ions follows:

`AgNO_3(aq.)+Br^-(aq.)\rightarrow AgBr(s)+NO_3^-(aq.)`

By Stoichiometry of the reaction:

1 mole of silver nitrate reacts with 1 mole of bromide ions.

So, 0.0174 moles of silver nitrate will react with =
`(1)/(1)* 0.0174=0.0174mol` of bromide ions.

Now, calculating the molarity of bromide ions by using equation 1, we get:

Moles of bromide ions = 0.0174 moles

Volume of solution = 50 mL = 0.05 L

Putting values in equation 1, we get:

`\text{Molarity of bromide ions}=(0.0174mol)/(0.05L)=0.348M`

Hence, the molarity of bromide ions is 0.348 M.