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Q: calcium carbonate (mg) dissolved per liter of rainwater ( No Answer,   3 Comments )
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 Subject: calcium carbonate (mg) dissolved per liter of rainwater Category: Science > Chemistry Asked by: hohum-ga List Price: \$200.00 Posted: 29 Jul 2006 01:43 PDT Expires: 28 Aug 2006 01:43 PDT Question ID: 750587
 ```The average pH of precipitation in rural areas is 5.10. Assuming that the pH is controlled by the carbonate system, i.e., no anthropogenic acid gases are present in the atmosphere, calculate the amount of calcium carbonate (mg) dissolved per liter of rainwater. pKa1=6.33, KCO2=3.388e-02, pCO2=3.162e-04```
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 ```Editors: I meant for this answer to only be a comment as it is somewhat speculative. Sorry- I hit the wrong key. Please remove it. Thanks!```
 ```I thought I knew some chemistry, but I don't understand. Try again offering less money, but more in layman's terms, if you are interested in less of an answer. My guess is the type of acid affects the amount disolved as well as the ph. Would a calculation for dilute H2SO4 serveyour purposes? Neil```
 ```There is somthing wrong with the data you have been given. In the absence of any dissolved CaCO3 (i.e., the pure CO2-H2O system) the pH of a solution in equilibrium with a partial pressure of CO2 = 3.162*10^-4 that one would calculate using the data you are given would be 5.65. Addition of CaCO3 would only serve to *increase* the pH (make the solution less acidic). The calculation of the pH for the CO2-H2O system goes as follows: A) [H+][OH-] = K_w = 10^-14 B) [H2CO3]/PCO2 = K_CO2 = 3.388*10^-2 (given, where K_CO3 is the Henry's Law coefficient for CO2 and water) C) [H+][HCO3-]/[H2CO3] = K_a1 = 10^-6.33 (given, where K_a1 is the first acidity constant for H2CO3) D) [H+][CO3-]/[HCO3--] = K_a2 = 10^-10.33 (this is something one can look up. It turns out it's not needed here, but I include it for completeness) E) [H+] - [OH-] - [HCO3-] - 2*[CO3--] = 0 (charge balance equation, i.e., the sum of the charges on the positive ions equals the sum of the charges on the negative ions) Begin by ignoring the second dissociation constant of H2CO3 (this is only important at high pHs, and we can show that this approximation is fine after we have run through the calculation). This assumption implies that [CO3--] = 0. Making this assumption, and using equation (A) to eliminate [OH-] in equation (E) gives: [H+] - 10^-14/[H+] = [HCO3-] Substitute this into equation (C), and use equation (B) to eliminate [H2CO3] to obtain: [H+]*([H+] - 10^-14/[H+])/(3.388*10^-2 * 3.162*10^-4) = 10^-6.33 [H+]^2 = 10^-6.33 * (3.388*10^-2 * 3.162*10^-4) + 10^-14 = 2.236 * 10^-6 M (only the positive root has any physical meaning) pH = -log[H+] = -log(2.236 * 10^-6) = 5.65. This is the pH of a solution in equilibrium with the specified P_CO2 and the given Henry's Law and dissociation constants. Using this value of [H+] in combination with equations (B) and (C), we can calculate [HCO3-]: [HCO3-] = 10^-6.33 * 3.388*10^-2 * 3.162*10^-4/(2.236 * 10^-6) = 2.241*10^-6 M To see that we were justified in neglecting the concentration of [CO3--], we can substutute these values for [H+] and [HCO3-] into equation (D) to see what the approximate concentration of [CO3--] would be: [CO3--] = 10^-10.33 * 2.241*10^-6/2.236*10^-6 = 4.69 * 10^-11 M, which is vanishingly small, so our approximation is ok. ------------------------------------------- For grins, let's crank through the calculations for the original problem, assuming everything is ok. We'll end up with an inconsistency, which will serve to show that there is a problem with the data. The chemical equilibria relevant to this system are: 1. H2O <-> H+ + OH- 2. CO2(g) + H2O(l) <-> H2CO2(aq) 3. H2CO3(aq) <-> H+ + HCO3- 4. HCO3- <-> H+ + CO3-- Because the problem does not specify that this solution is in equilibrium with solid CaCO3, the dissolution reaction of calcite (CaCO3) does not add any additional equilibrium constraints relative to the problem solved above. For each of these reactions, we can write an equilibrium constant: 1a. [H+][OH-] = K_w = 10^-14 2a. [H2CO3]/PCO2 = K_CO2 = 3.388*10^-2 3a. [H+][HCO3-]/[H2CO3] = K_a1 = 10^-6.33 4a. [H+][CO3-]/[HCO3--] = K_a2 = 10^-10.33 Plus a charge-balance equation: 5a. [H+] + 2*[Ca++] = [OH-] + [HCO3-] + 2*[CO3--] Note that now we are admitting the possibility that there are Ca++ ions in solution. We are told that the pH = 5.1, so [H+] = 10^-5.1 M. This immediately (from 1a) gives that [OH-] = 10^-14/10^-5.1 = 10^-8.9 M. Equation 2a and the fact that PCO2 = 3.162*10^-4 gives us that [H2CO3] = 3.162*10^-4 * 3.388*10^-2 = 1.071*10^-5 M Using this result, and the value for [H+] found above in equation 3a gets us [HCO3-]: [HCO3-] = 10^-6.33*[H2CO3]/[H+] = 10^-6.33 * 1.071*10^-5/10^-5.1 = 5.011*10^-12 M Because we are at a pH of 5.1, we can neglect the second dissociation constant of H2CO3, just as we did above. The concentration of CO3-- at acidic pH's is vanishingly small. (To see this, plug in values for [H+] and [HCO3-] in equation 4a above.) We now know the concentrations of all the charged species in this solution except for [Ca++]. TO find the concentration of this ion, plug the other values into the charge-balance equation (equation 5a) and solve for [Ca++]: [Ca++] = 1/2*([OH-] + 2*[CO3--] + [HCO3] - [H]) = -3.971 * 10^-6 M, a negative quantity, which doesn't make any sense physically!```