Solubility product of silver bromide is 5.0 x 10-13. The quantity of p...
For precipitation
[Ag+][Br-] > Ksp[AgBr]
[Br-]min = (5 x 10-13)/0.05 = 10-11 M
Mass of potassium bromide needed = 10-11 x 120
= 1.2 x 10-9 g
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Solubility product of silver bromide is 5.0 x 10-13. The quantity of p...
To start the precipitation of AgBr, we need to add an appropriate amount of potassium bromide (KBr) to a solution of silver nitrate (AgNO3). The solubility product of silver bromide (AgBr) is given as 5.0 x 10^-13.
Let's calculate the number of moles of Ag+ ions in 1 liter of 0.05 M silver nitrate solution.
Number of moles of Ag+ = Molarity x Volume
= 0.05 mol/L x 1 L
= 0.05 mol
Since the solubility product of AgBr is given as 5.0 x 10^-13, we can write the equilibrium expression for the precipitation of AgBr as follows:
Ag+ + Br- → AgBr
According to the stoichiometry of the reaction, 1 mole of Ag+ reacts with 1 mole of Br-. Therefore, the concentration of Br- ions in the solution will be equal to the concentration of Ag+ ions, which is 0.05 mol/L.
Now, let's calculate the number of moles of KBr required to provide the necessary Br- ions.
Number of moles of KBr = Moles of Br- ions required
= 0.05 mol
The molar mass of KBr is given as 120 g/mol. Using this information, we can calculate the mass of KBr required.
Mass of KBr = Number of moles x Molar mass
= 0.05 mol x 120 g/mol
= 6 g
However, we need to convert this mass to grams because the options are given in grams. The conversion factor between grams and micrograms is 1 g = 10^6 µg.
Therefore, the mass of KBr required in micrograms is:
Mass of KBr = 6 g x 10^6 µg/g
= 6 x 10^6 µg
Comparing this with the given options, we can see that option C is the closest value:
Option C) 1.2 x 10^-9 g
Hence, the correct answer is option C) 1.2 x 10^-9 g.
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