Question

Potassium tetraiodomercurate (K2HgI4) is aqueous solution goes by several names, including Nesslers reagent and Mayers reagent. It is an effective test for organic alkaloids and ammonia. A typical batch is made by dissolving 1.36 g of mercuric chloride and 5.00 g of potassium iodide in 20.0 mL of water. (a) Calculate the molal concentration of K2HgI4 in the solution. (b) Calculate the cryoscopic constant for water. (c) This solution has a freezing point of 271.74K. Calculate the molality based on this observation. Explain what is happening.

Answer 1

(a): Given the mass of HgCl2 = 1.36 g

Moles of HgCl2 taken = mass / molar mass = 1.36 g / 271.52 g/mol = 0.005009 mol HgCl2

Given the mass of KI = 5.00 g

Moles of KI taken = mass / molar mass = 5.0 g / 166.00 g/mol = 0.03012 mol KI

The balanced equation for the formation of K2HgI4 is

HgCl2 + 4KI ----- > K2HgI4 + 2KCl

1 mol, 4 mol ------ 1 mol, --- 2 mol

1 mol of HgCl2 reacts with 4 mol of KI

Hence 0.005009 mol HgCl2 that will react with the moles of KI

= 0.005009 mol HgCl2 x ( 4 mol KI / 1 mol HgCl2) = 0.020036 mol KI

Hence HgCl2 is used up completely and acts as limiting reactant and it will decide the amount of K2HgI4 formed.

Hence moles of K2HgI4 formed =  0.005009 mol HgCl2 x ( 1 mol K2HgI4 / 1 mol HgCl2) = 0.005009 mol K2HgI4.

Volume of the solvent(water) , V = 20.0 mL

mass of water = 20.0 mLx 1 g/mL = 20 g = 0.020 Kg

Hence molal concentration of K2HgI4 =  0.005009 mol K2HgI4 / 0.020 Kg = 0.2504 molal (answer)

(b): Cryoscopic constant of water, Kf = 1.86 DegC.m^-1

Given freezing point, Tf = 271.74 K

Tf = Kf x m

=> m = Tf / Kf = (273.14 - 271.74) / 1.86 DegC.m^-1 = 0.753 m (answer)

Since K2HgI4 is completely dissociated into 2K+(aq) and HgI4-(aq), also KI is dissociated into K+(aq) and I-(aq), the observed molality is very high.