Question

A 25-ml sample of river water was titrated with 0.0010 m k2cr2o7 and required 8.3 ml to reach the end point. what is the chemical oxygen demand, in milligrams of o2 per liter, of the sample?

Answer 1

First let us calculate the moles of K2Cr2O7 that was supplied.

moles K2Cr2O7 = 0.0010 M * 0.0083 L = 8.3x10^-6 mol

 

From the chemical formula itself, we see that there are 7 O for every mole of K2Cr2O7 or 3.5 O2. Therefore:

moles O2 = 8.3x10^-6 mol K2Cr2O7 * (3.5 mol O2 / 1 mol K2Cr2O7)

moles O2 = 2.905x10^-5 mol O2

 

Calculating for the mass of O2 in mg:

mass O2 = 2.905x10^-5 mol O2 * (32 g / mol) * (1000 mg / g)

mass O2 = 0.9296 mg

 

Therefore the chemical oxygen demand (COD) is:

COD = 0.9296 mg / (0.025 L)

COD = 37.184 mg/L