Post on 02-Feb-2016
description
What kind of lake?
A calcareous lake!
= 2[Ca2+], from:
6.35
(So [H+] = 5.26 x 10-9)
Now add H+ (remember, [H+]0 = 5.26 x 10-
9):1) Enough to change [H+] by 100X if no HCO3
-:[HCO3
-]=1.02x10-3 - 5.21x10-7=1.02 x 10-3, [H2CO3]=1.25x10-5 and pH = 8.3; pH = 0.0
2) Enough to change [H+] by 1,000X “ “ “:[HCO3
-]=1.02x10-3 - 5.26x10-6=1.015 x 10-3, [H2CO3]=1.72x10-5 and pH = 8.1; pH = -0.2
3) Enough to change [H+] by 10,000X “ “ “:[HCO3
-]=1.02x10-3 - 5.26x10-5=9.67x10-4, [H2CO3]=6.46x10-5 and pH = 7.5; pH = -0.8
4) Enough to change [H+] by 100,000X “ “ “:[HCO3
-]=1.02x10-3 - 5.26x10-4=4.94x10-4, [H2CO3]=5.38x10-4 and pH = 6.3; pH = -2.0
H2CO3 H+ + HCO3-
]CO[H][HCO
log pK pH32
-3
a
5-
3-
10 1.2
10 1.02 log 6.35 pH
In plain H2O: pH=-log(6.21 x 10-7) = 6.2; =-0.8!
In plain H2O: pH=-log(5.36 x 10-6) = 5.3; =-1.7!
In plain H2O: pH=-log(5.27 x 10-5) = 4.3; =-2.7!
In plain H2O: pH=-log(5.26 x 10-4) = 3.3; =-3.7!
H2CO3 H+ + HCO3-
]CO[H][HCO
log pK pH32
-3
a
5-
3-
10 1.2
10 1.02 log 6.35 pH
Now add OH- (remember, [OH-]0 = 1.90 x 10-
6):1) Enough to change [OH-] by 2X if no H2CO3:[H2CO3]=1.2x10-5 - 1.90 x 10-6=1.01x10-5, [HCO3
-]=1.022x10-3 and pH = 8.4; pH= 0.1
2) Enough to change [OH-] by 5X if no H2CO3:[H2CO3]=1.2x10-5 - 7.6x10-6=4.40 x 10-6,[HCO3
-]=1.028x10-3 and pH = 8.7; pH = 0.4
3) Enough to change [H+] by 7X “ “ “:[H2CO3]=1.2x10-5 - 1.14x10-5=6.00x10-7 (5%)[HCO3
-]=1.031x10-3 and pH = 9.6; pH = 1.3
In plain H2O: pOH=-log(2.00 x 10-6) = 5.7 pH = 14 - pOH = 8.3; = 1.3!
In plain H2O: pOH=-log(7.7 x 10-6) = 5.1 pH = 14 - pOH = 8.9; = 1.9!
In plain H2O: pOH=-log(1.15 x 10-5) = 4.9 pH = 14 - pOH = 9.1; = 2.1!