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!