Question

A solution was prepared by dissolving 0.1937 g of HgO (table) in 20 mL of water containing 4 g of KBr. Titration with HCl required 17.98 mL to reach a phenolphthalein endpoint. Calculate the molarity of the HCl.

Answer 1

Given the mass of HgO = 0.1937 g

Hence moles of HgO = mass / molar mass of HgO = 0.1937 g / 216.60 g/mol = 8.943x10^-4 mol

mass of KBr = 4.0 g

Hence moles of KBr = mass / molar mass of KBr = 4.0 g / 119.0 g/mol = 0.03361 mol

The balanced equation for the reaction of HgO with KBr is

HgO + 4 Br⁻  +  H₂O ----> HgBr₄^2- +  2 OH⁻

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

From the above reaction it is clear that 1 mol of HgO reacts with 4 mol of Br^- to produce 2 mol of OH-ion.

Hence 8.943x10^-4 mol of HgO that will react with the moles of Br-

= 8.943x10^-4 mol HgO x ( 4 mol Br^- / 1 mol HgO) = 0.0035772 mol Br-, which is less than the moles of Br- initially taken.

Hence HgO is the limiting reactant and decides the moles of product formed.

Moles of OH- formed by 8.943x10^-4 mol HgO

= 8.943x10^-4 mol HgO x ( 2 mol OH- / 1 mol HgO)  = 0.0017886 mol OH-

Now the chmical equation for the reaction of OH- and HCl is

HCl + OH- ---- > H2O + Cl-

Hence the moles of HCl required for the neutralization of OH- = 0.0017886 mol OH-

Volume of HCl taken = 17.98 mL = 0.01798 L. Let the concentration be 'M'

Moles of HCl taken = MxV = M mol/Lx 0.01798 L = 0.01798M mol

=> 0.01798M mol = 0.0017886 mol OH-

=> M =  0.0017886 mol / 0.01798 = 0.0995 M (answer)