Problem 2 (16 points) Given the following: Fe2-202 20-Fe\" (OH), +2H. K 625 x 10
ID: 637126 • Letter: P
Question
Problem 2 (16 points) Given the following: Fe2-202 20-Fe" (OH), +2H. K 625 x 10O L min mol Initial dissolved oxygen: DO 268mM Initial iron concentration : Fe2+-5.58 mg/L pH 6.0 The rate law for the loss of iron,-=-k [FeTomT [DO] a) Calculate the rate of production/loss of Fe2 b) For a final Fe concentration of 0.3 mg/L, determine the concentration 6 marks] and rate of production/loss of DO, Fe(OH)3, alkalinity, and acid. Assume a constant pH. Use a stoichiometric table to obtain the values. [10 marks] [This question was in the 2017 mid-semester test]Explanation / Answer
(a) The rate law for loss of Fe2+ = -k[Fe2+][OH-]2[DO]
GIVEN DATA
Fe2+=5.58mg/l
DO =.268mM
pH=6
as we know
pH+pOH =14
pOH=14-6=8
pOH =-log[OH-]
[OH-]=10-8
PUTTING ALL VALUES IN RATE LAW EQUATION
r = -6.25*1016 *.268 *5.58 *10-8
= -9.34*108 divide it by 1000 because DO is given in mM so to convert it to mole
= -9.34*105 mg/min