Astudent 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?

The molarity of bromide ions is 0.348 M.

Explanation:

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

      .....(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:

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

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 = 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:

Hence, the molarity of bromide ions is 0.348 M.

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