AT&T; LTE 3:59 PM blackboard.sc.edu Answer on a separate page, showing your calc

Question

AT&T; LTE 3:59 PM blackboard.sc.edu Answer on a separate page, showing your calculations and explaining your assumptions, NEATLY For the following problems use the following K,= 0.034 M/atm = 10.47 M/atm pk = 5.85 (where K, = [H] x [HCO, ] / [COL pK, = 8.97 (where K' = [H1 x [CO,'] / [HCO, ] 1. What is the pH of pure water in equilibrium with 400 ppm CO, (-4.0 x 10' atm 10"atm) a. Why can you assume charge balance only involves H and HCO,? b. What is the pH of the pure water? 2. What is the pH of seawater with [CO] = 1.02 x 10' M in equilibrium with 316 ppm Co,? 3. What happens to the pH when CO, levels increase to 400 ppm? Extra credit: What is the percent increase in acidity? The key to answering this is understanding the nature of the pH scale (it is logarithmic, not linear)

Explanation / Answer

Solution ;-

From henry law

C = KH × P

= 0.034(M/atm) × 4 × 10^-4 atm

= 1.36 × 10^-3M

CO2 cconcentration in solution = 1.36 × 10^-3M

Concentration of H2CO3 = 1.36 × 10^-3M

  H2CO3(aq)< ---------> HCO3- (aq) + H+(aq)

   Ka= [HCO3-] [ H+]/ [ H2CO3]

1.413 ×10^-6 = X^2/1.36 × 10^-3

   X = 4.385 × 10^-5

[H+] = 4.385 × 10^-5M

pH = 4.35

a) pka2 = 8.97

Ka2 = 1.072 × 10^-9

Ka2 = [H+] [ CO3^2-]/ [HCO3-]

1.072 × 10^-9 = X^2/4.385× 10^-5

X = 2.17 × 10-7

[H+ ] = 2.17 × 10^-7M

Adding this hydrogen ion concentration to hydrogen ion concentration from ka1 equilibrium is not changing considerably , so pH change is not considerable. Therefore, we can assume charge balance involve H+ and HCO3- only.

b) pH of pure water = 7.0

2. Ka2 = 1.072 × 10^-9M

1.072 × 10^-9 = (1.02× 10^-4)^2/[HCO3-]

[HCO3-] = 1.04×10^-8/1.072 × 10^-9

= 9.70M

[ H+ ] = 9.70 M

pH = 0.99

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