Question

A water sample contains 150 mg/L ethylene glycol (C2H6O2) and 200 mg/L phenol (C6H6O). Calculate the COD and TOC.

Answer 1

COD = Chemical oxygen demand

TOC=Total organic carbon .

1) COD= A) ethylene glycol( C2H6O2)       (12×2+1×6+32)=62

C2H6O2+2.5 O2-→2CO2+3H2O

COD =2.5×32/62×150mg/l= 194mg/l

B) Phenol (C6H6O)      (12×6+1×6+16) = 94

COD= 7×32/94×200mg/l = 476.59mg/l

2) TOC = A) ethylene glycol( C2H6O2)       (12×2+1×6+32)=62g/mol

Z×12g/mol         2×12=24g/mol                  

24g/mol/62g/ml=0.387  

150mg/l×0.387 = 58.05mg/l

B) TOC = Phenol (C6H6O)      (12×6+1×6+16) = 94g/mol

Carbon mass per mol of satandard compound Z×12 g/mol

6×12gm/mol = 72g/mol

Carbon mass portion of standard compound mass

72gm/mol/94gm/mol= 0.766    phenol mass concentration of standard solution

200gm/l  

0.766×200mg/l = 153.2mg/l