What kind of lake? A calcareous lake!.

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Presentation transcript:

What kind of lake? A calcareous lake!

= 2[Ca2+], from:

6.35 (So [H+] = 5.26 x 10-9)

H2CO3  H+ + HCO3- 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 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- 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!