4. Consider a pool (25°C) in which the water has a sulphate concentration [SO₄²] of 28 mm and is in equilibrium with the minerals calcite (CaCO3) and anhydrite (CaSO4). a. Calcula... 4. Consider a pool (25°C) in which the water has a sulphate concentration [SO₄²] of 28 mm and is in equilibrium with the minerals calcite (CaCO3) and anhydrite (CaSO4). a. Calculate [Ca²⁺] concentration. b. Calculate [CO₃²⁻] concentration. c. Calculate the pH. 5. Surface water in full exchange with the atmosphere with a partial pressure of CO₂ (pCO₂) of 10⁻³·⁵ atm at 25°C. The dissociation constants (pK₁ and pK₂) are 6.35 and 10.33, respectively and Henry's constant (KH) = 10⁻¹·⁴⁷. a. Calculate the total inorganic carbon concentration (CT) in mM at a pH of 8.3 and 7.6. b. Calculate the total alkalinity in mM at a pH of 8.3 and 7.6. c. Calculate the carbonate alkalinity at a pH of 8. d. Calculate the Ca²⁺ concentration in mM if the water is in equilibrium with the mineral calcite (CaCO3) at a pH of 7.8. 6. Give 3 reasons why calcite (CaCO3) dissolves with increasing depth in the ocean. Read more

Steps to Solve

Problem 4

1. Calculate [Ca²⁺] concentration

Since the pool is in equilibrium with both calcite ($CaCO_3$) and anhydrite ($CaSO_4$), the concentration of $Ca^{2+}$ is determined by the dissolution of anhydrite, because anhydrite is more soluble than calcite. Given that the sulphate concentration $[SO_4^{2-}]$ is 28 mM, this is equal to the $[Ca^{2+}]$. $$[Ca^{2+}] = [SO_4^{2-}] = 28 \text{ mM}$$

2. Calculate [CO₃²⁻] concentration

The pool is in equilibrium with calcite, so we can use the solubility product ($K_{sp}$) of calcite to calculate $[CO_3^{2-}]$. The $K_{sp}$ for calcite is approximately $4.5 \times 10^{-9}$. $$K_{sp} = [Ca^{2+}][CO_3^{2-}]$$ Rearranging for $[CO_3^{2-}]$: $$[CO_3^{2-}] = \frac{K_{sp}}{[Ca^{2+}]} = \frac{4.5 \times 10^{-9}}{28 \times 10^{-3}}$$ $$[CO_3^{2-}] = \frac{4.5 \times 10^{-9}}{0.028} \approx 1.61 \times 10^{-7} \text{ M}$$

3. Calculate the pH

To calculate the pH, we need to consider the equilibrium of carbonate ions with water. The relevant reaction is: $$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$$ We can calculate the $K_b$ for this reaction using the relationship $K_w = K_a \times K_b$, where $K_w = 1.0 \times 10^{-14}$. We need to use the $K_a$ of $HCO_3^-$, which is the second dissociation constant of carbonic acid, $K_2 = 10^{-10.33}$. $$K_b = \frac{K_w}{K_2} = \frac{1.0 \times 10^{-14}}{10^{-10.33}} = 10^{-3.67} \approx 2.14 \times 10^{-4}$$ Now, we can set up an ICE table: Initial: $[CO_3^{2-}] = 1.61 \times 10^{-7}$, $[OH^-] = 0$ Change: $[CO_3^{2-}] = -x$, $[OH^-] = +x$ Equilibrium: $[CO_3^{2-}] = (1.61 \times 10^{-7}) - x$, $[OH^-] = x$ $$K_b = \frac{[OH^-][HCO_3^-]}{[CO_3^{2-}]} = \frac{x^2}{1.61 \times 10^{-7} - x} \approx \frac{x^2}{1.61 \times 10^{-7}} = 2.14 \times 10^{-4}$$ Solving for $x$: $$x^2 = (2.14 \times 10^{-4}) \times (1.61 \times 10^{-7}) \approx 3.44 \times 10^{-11}$$ $$x = \sqrt{3.44 \times 10^{-11}} \approx 5.86 \times 10^{-6} \text{ M}$$ So, $[OH^-] = 5.86 \times 10^{-6}$.

Now calculate the pOH: $$pOH = -log[OH^-] = -log(5.86 \times 10^{-6}) \approx 5.23$$ Finally, calculate the pH: $$pH = 14 - pOH = 14 - 5.23 \approx 8.77$$

Problem 5

1. Calculate the total inorganic carbon concentration ($C_T$) in mM at a pH of 8.3 and 7.6

The total inorganic carbon concentration ($C_T$) can be calculated using the following formula: $$C_T = [H_2CO_3^] + [HCO_3^-] + [CO_3^{2-}]$$ Where $[H_2CO_3^] = [CO_2(aq)] = K_H \times pCO_2$. Given $K_H = 10^{-1.47}$ and $pCO_2 = 10^{-3.5}$ atm: $$[H_2CO_3^] = 10^{-1.47} \times 10^{-3.5} = 10^{-4.97} \text{ M}$$ We can also express the other species in terms of $H_2CO_3^$: $$[HCO_3^-] = [H_2CO_3^] \times 10^{pH - pK_1}$$ $$[CO_3^{2-}] = [HCO_3^-] \times 10^{pH - pK_2} = [H_2CO_3^] \times 10^{(pH - pK_1) + (pH - pK_2)}$$ Now calculate for pH = 8.3: $$[HCO_3^-] = 10^{-4.97} \times 10^{8.3 - 6.35} = 10^{-4.97} \times 10^{1.95} = 10^{-3.02} \text{ M}$$ $$[CO_3^{2-}] = 10^{-3.02} \times 10^{8.3 - 10.33} = 10^{-3.02} \times 10^{-2.03} = 10^{-5.05} \text{ M}$$ $$C_T = 10^{-4.97} + 10^{-3.02} + 10^{-5.05} \approx 1.07 \times 10^{-3} \text{ M} = 1.07 \text{ mM}$$

Now calculate for pH = 7.6: $$[HCO_3^-] = 10^{-4.97} \times 10^{7.6 - 6.35} = 10^{-4.97} \times 10^{1.25} = 10^{-3.72} \text{ M}$$ $$[CO_3^{2-}] = 10^{-3.72} \times 10^{7.6 - 10.33} = 10^{-3.72} \times 10^{-2.73} = 10^{-6.45} \text{ M}$$ $$C_T = 10^{-4.97} + 10^{-3.72} + 10^{-6.45} \approx 2.09 \times 10^{-4} \text{ M} = 0.209 \text{ mM}$$

2. Calculate the total alkalinity in mM at a pH of 8.3 and 7.6

The total alkalinity is given by: $$TA = [HCO_3^-] + 2[CO_3^{2-}] + [OH^-] - [H^+]$$ At pH = 8.3: $[H^+] = 10^{-8.3}$ M, $[OH^-] = 10^{-(14 - 8.3)} = 10^{-5.7}$ M $$TA = 10^{-3.02} + 2 \times 10^{-5.05} + 10^{-5.7} - 10^{-8.3} \approx 9.55 \times 10^{-4} \text{ M} = 0.955 \text{ mM}$$ At pH = 7.6: $[H^+] = 10^{-7.6}$ M, $[OH^-] = 10^{-(14 - 7.6)} = 10^{-6.4}$ M $$TA = 10^{-3.72} + 2 \times 10^{-6.45} + 10^{-6.4} - 10^{-7.6} \approx 1.90 \times 10^{-4} \text{ M} = 0.190 \text{ mM}$$

3. Calculate the carbonate alkalinity at a pH of 8

Carbonate alkalinity ($CA$) is given by: $$CA = [HCO_3^-] + 2[CO_3^{2-}]$$ First, calculate $[H_2CO_3^]$: $$[H_2CO_3^] = K_H \times pCO_2 = 10^{-1.47} \times 10^{-3.5} = 10^{-4.97} \text{ M}$$ Now calculate $[HCO_3^-]$ and $[CO_3^{2-}]$: $$[HCO_3^-] = [H_2CO_3^*] \times 10^{pH - pK_1} = 10^{-4.97} \times 10^{8 - 6.35} = 10^{-4.97} \times 10^{1.65} = 10^{-3.32} \text{ M}$$ $$[CO_3^{2-}] = [HCO_3^-] \times 10^{pH - pK_2} = 10^{-3.32} \times 10^{8 - 10.33} = 10^{-3.32} \times 10^{-2.33} = 10^{-5.65} \text{ M}$$ $$CA = 10^{-3.32} + 2 \times 10^{-5.65} \approx 4.79 \times 10^{-4} \text{ M} = 0.479 \text{ mM}$$

4. Calculate the $Ca^{2+}$ concentration in mM if the water is in equilibrium with the mineral calcite ($CaCO_3$) at a pH of 7.8

The $K_{sp}$ for calcite is approximately $4.5 \times 10^{-9}$. We know that $K_{sp} = [Ca^{2+}][CO_3^{2-}]$. We need to calculate $[CO_3^{2-}]$ at pH = 7.8. First calculate $[H_2CO_3^]$ $$[H_2CO_3^] = K_H \times pCO_2 = 10^{-1.47} \times 10^{-3.5} = 10^{-4.97} \text{ M}$$ Then calculate $[HCO_3^-]$ $$[HCO_3^-] = [H_2CO_3^*] \times 10^{pH - pK_1} = 10^{-4.97} \times 10^{7.8 - 6.35} = 10^{-4.97} \times 10^{1.45} = 10^{-3.52} \text{ M}$$ Then calculate $[CO_3^{2-}]$ $$[CO_3^{2-}] = [HCO_3^-] \times 10^{pH - pK_2} = 10^{-3.52} \times 10^{7.8 - 10.33} = 10^{-3.52} \times 10^{-2.53} = 10^{-6.05} \text{ M}$$ Now calculate $[Ca^{2+}]$ $$[Ca^{2+}] = \frac{K_{sp}}{[CO_3^{2-}]} = \frac{4.5 \times 10^{-9}}{10^{-6.05}} = 4.5 \times 10^{-2.95} = 1.13 \times 10^{-3} \text{ M} = 1.13 \text{ mM}$$

Problem 6

1. Give 3 reasons why calcite ($CaCO_3$) dissolves with increasing depth in the ocean.

   • Increased Pressure: The solubility of calcite increases with increasing pressure. At greater depths, the higher pressure favors the dissolution of $CaCO_3$ to reduce the overall volume.
   • Decreased Temperature: The solubility of calcite generally increases as temperature decreases. Deep ocean water is colder than surface water, which promotes dissolution.
   • Increased $CO_2$ Concentration (Lower pH): At greater depths, the concentration of carbon dioxide ($CO_2$) increases due to the decomposition of organic matter and respiration by marine organisms. Increased $CO_2$ leads to a lower pH (more acidic conditions), which enhances the dissolution of calcite according to the reaction: $CaCO_3(s) + CO_2(aq) + H_2O(l) \rightleftharpoons Ca^{2+}(aq) + 2HCO_3^-(aq)$