Suppose of potassium bromide is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the potassium bromide is dissolved in it. Round your answer to significant digits.

Suppose 0.549g of potassium bromide is dissolved in 100.mL of a 19.0mM aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the potassium bromide is dissolved in it. Round your answer to 3 significant digits.

We first assume the reaction goes to completion and that the volume of the reaction does not change.

The balanced stoichiometric equation is

KBr + AgNO₃ → KNO₃ + AgBr

First of, we need to check which reactant is in excess and which one is the limiting reagent (The one that is completely used up in the reaction). To do this, we find the number moles of each reactant at the start of the reaction.

For KBr

Number of moles = (Mass)/(Molar Mass)

Mass of Potassium Bromide = 0.549 g

Molar mass of Potassium Bromide = 119.002 g/mol

Number of moles = (0.549/119.002) = 0.00461 moles.

For AgNO₃,

Number of moles = (Concentration in mol/L) × (Volume in L)

Concentration of Silver Nitrate in mol/L = (19/1000) M = 0.019 M

Volume in L = (100/1000) = 0.100 L

Number of moles = 0.019 × 0.100 = 0.0019 moles

From the stoichiometric balance of the reaction,

1 mole of KBr reacts with 1 mole of AgNO₃

Hence, it is evident that it is AgNO₃ that is the limiting reagent; the chemical specie that is used up in the process of the reaction and it determines the amount of other reactants required & products formed.

1 mole of AgNO₃ gives 1 moles of AgBr

0.0019 moles of AgNO₃ will give 1 × 0.0019 moles of AgBr; 0.0019 moles of AgBr

Note that 1 mole of AgBr contains 1 mole of Br⁻ ion.

0.0019 mole of AgBr contains 0.0019 mole of Br⁻ ion.

Note that there is an excess of KBr because 0.00461 mole was available pre-reaction and according to the stoichiometric balance of the reaction, 0.0019 mole of KBr was used up in the reaction.

Excess KBr = 0.00461 - 0.0019 = 0.00271 moles.

1 mole of KBr contains 1 mole of Br⁻

0.00271 moles of excess KBr will contain 0.00271 moles of Br⁻

Total number of moles Br⁻ ion available at the end of the reaction = 0.0019 + 0.00271 = 0.00461 moles

Molarity = (Number of moles) ÷ (Volume in L)

Number of moles of Br⁻ left in the reaction mixture = 0.00461 moles

Vol in L of the mixture = 0.100 L (the volume of the reacting mixture doesn't change)

Molarity of Br⁻ = (0.00461/0.1) = 0.0461 M

Hope this Helps!!!!