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
