Question

Suppose of potassium iodide is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

0.0983 M

Explanation:

First, we need to find the formulas of the reactants. Potassium forms the ion K⁺, and iodide is the ion I⁻, thus potassium iodide is KI. Silver forms the ion Ag⁺, and nitrate is the ion NO₃⁻, thus silver nitrate is AgNO₃. In the reaction, the cations will be replaced:

KI(aq) + AgNO₃(aq) → KNO₃(aq) + AgI(s)

AgI is an insoluble salt, so it will precipitate, and all nitrates are soluble, thus KNO₃ will be in the ionic form: K⁺ and NO₃⁻. 1 mol of KNO₃ = 1 mol of K⁺.

The molar mass of KI is 166 g/mol, thus the number of moles that is added is:

nKI = mass/molar mass

nKI = 5.71/166 = 0.0344 mol

And the number of moles of AgNO₃ is given as 64mM = 0.064 mol. Because the stoichiometry is 1:1, AgNO₃ is in excess, thus, all the KI will react and form 0.0344 mol of KNO₃. So, nK⁺ = 0.0344 mol. The molarity is the number of moles divided by the volume (350 mL = 0.350 L):

0.0344/0.350 = 0.0983 M