1. What kind of lake?
A calcareous lake!

13. = 2[Ca2+], from:

14. 6.35
(So [H+] = 5.26 x 10-9)

15. 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-]=1.02x x10-7=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-]=1.02x x10-6=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-]=1.02x x10-5=9.67x10-4, [H2CO3]=6.46x10-5 and pH = 7.5; pH = -0.8
4) Enough to change [H+] by 100,000X “ “ “:
[HCO3-]=1.02x x10-4=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!

16. 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]=1.2x 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]=1.2x x10-6=4.40 x 10-6,
[HCO3-]=1.028x10-3 and pH = 8.7; pH = 0.4
3) Enough to change [H+] by 7X “ “ “:
[H2CO3]=1.2x x10-5=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!