Question

What is the level of saturation of lake water containing 10.8 mg/L dissolved oxygen at 14^o C. Note that K_H (14^o C)= 1.53*10^-3 mol/L*atm.

I know that [O₂]_eq=K_H*P_O2 but I am unsure how I should approach this problem otherwise.

Answer 1

equilibrium oxygen concentration at non standard temperature and pressure = [O₂]eq

%saturation = [dissolved oxygen (mg/L) / [O₂]eq ]*100 ................ equation 1

Dissolved oxygen = 10.8 mg/L ( as given in question)

[O2]eq=KH*PO2 ..............................equation2

First we will calculate pO₂ , which is equal to product of mole fraction of oxygen and total atmospheric pressure ....equation3

standard atmospheric pressure(P1) =1 atm

standard atmospheric temerature (T1) =273k

final temperature(T2)= 287k

final pressure (P2)=

P₁T₂=P₂T₁

P₂= (287*1) /273

P₂₌  1.0513 atm

mole fraction of oxygen is 21%  

according to equation 3

pO₂=  ​1.0513 * 0.21

= 0.2207 atm

now put value of pO₂ in equation 2

[O₂]eq=KH*PO2

= 1.53*10^-3(molL^-1atm-1) * 0.2207(atm)

=1.53*32(mgL-1atm^-1) *0.2207(atm)

[O₂]eq = 10.8 mgL-1

%saturation = [dissolved oxygen (mg/L) / [O₂]eq ]*100

= (10.8/10.8) * 100

= 100

ANSWER = level of saturation of lake water IS 100%