Time, min [Fe(ll)], mg/L In [Fe [Fe(II)], mg/L In [Fe(II)] [Fe(ll)], mg/L n [Fe(

Question

Time, min [Fe(ll)], mg/L In [Fe [Fe(II)], mg/L In [Fe(II)] [Fe(ll)], mg/L n [Fe(ll] pH- 6.0 pH- 6.0 pH 6.25 pH 6.25 pH 6.5 pH 6.5 5.858 350.0 5.858 350.0 5.858 321.8 5.831 5.774 340.8 5.850 347.1 5.805 295.9 5.841 331.9 344.2 10 5.778 272.0 5.606 323.2 5.833 341.3 15 314.7 5,752 250.1 5.522 5.824 338.4 5.725 230.0 5.438 306.5 5.816 335.6 25 5.699 298.4 211.4 5.354 5.808 332.8 178.7 5.791 283.0 5.645 5.186 327.3 40 268.4 5.592 5.774 151.1 5.018 321.8 5.757 254.5 127.7 316.4 4.850 5.740 108.0 241.3 5.486 311.2 4.682 5.724 5.433 306.0 228.8 91.3 4,514 217.0 300.9 5.707 5.380 77.2 4.346 295.9 5.690 205.8 5.327 65.2 100 4.178 5.274 110 5.673 195.1 55.1 4.010 120 286.1 5.656 185.0 46.6 3.842 pH 6.00 -o phi 6.25 pH 6.50 pH 6.00 OpH 6.25 A pH 6.50 350.0 6.0 slope -0.00168 300.0 5.5 slope -0.00531 250.0 5.0 200.0 slope -0.01680 150.00 4.0 E 100.0 3.5 50.0 3.0 0 20 40 60 80 100 120 0.0 0 20 40 60 80 100 120 Time, min Time, min

Explanation / Answer

1.

pH=6

pOH=8

So [OH-]=10-pOH=10-8 M

Initial

[Fe2+]=350 mg/lit=350*10-3 gm/lit/55.8 gm/mol=6.27*10-3 M

So using equation

slope=d[Fe]/dt=-0.00168

So Using formula

-0.00168=-k*6.27*10-3*(10-8)2*0.21

k=1.276*1016 M-2.atm-1.min-1

k'=1.276*1015*10-16*0.21=0.2678 min-1

2.

pH=6.25

pOH=7.75

So [OH-]=10-pOH=10-7.75=1.778*10-8 M

Initial

[Fe2+]=350 mg/lit=350*10-3 gm/lit/55.8 gm/mol=6.27*10-3 M

So using equation

slope=d[Fe]/dt=-0.00531

So Using formula

-0.00531=-k*6.27*10-3*(1.778*10-8)2*0.21

k=1.275*1016 M-2.atm-1.min-1

k'=1.275*1016*(1.778*10-8)2*0.21=0.8466 min-1

3.

pH=6.5

pOH=7.5

So [OH-]=10-pOH=10-7.5=3.16*10-8 M

Initial

[Fe2+]=350 mg/lit=350*10-3 gm/lit/55.8 gm/mol=6.27*10-3 M

So using equation

slope=d[Fe]/dt=-0.0168

So Using formula

-0.0168=-k*6.27*10-3*(3.16*10-8)2*0.21

k=1.276*1016 M-2.atm-1.min-1

k'=1.276*1016*(3.16*10-8)2*0.21=2.675 min-1

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