The pH of a particular raindrop is 5.6.
(a) Assuming the major species in the raindrop are H2CO3(aq), HCO3-(aq) and CO32- (aq), Calculate the concentrations of these species in the raindrop, assuming the total carbonate concentration is 1.0 * 10-5 M. The appropriate Ka values are given in Table 16.3.
(b) What experiments could you do to test the hypothesis that the rain also contains sulfur-containing species that contribute to its pH? Assume you have a large sample of rain to test.

Question

The pH of a particular raindrop is 5.6.

(a) Assuming the major species in the raindrop are H2CO3(aq), HCO3-(aq) and CO32- (aq), Calculate the concentrations of these species in the raindrop, assuming the total carbonate concentration is 1.0 * 10-5 M. The appropriate Ka values are given in Table 16.3.

(b) What experiments could you do to test the hypothesis that the rain also contains sulfur-containing species that contribute to its pH? Assume you have a large sample of rain to test.

Accepted Answer

Hello everyone today. With the following problem. A normal and clean rainwater has been found to have a ph of 5.8 which is attributed to the presence of carbonic acid. Carbonic acid formed from dissolved carbon dioxide gas and water. Calculate the concentration of carbonic acid, carbonate, bicarbonate and carbonate. Assuming that the total carbonate concentration is one times 10 to the negative fifth molar. And that the given acid ionization for the 1st and 2nd equation values are 4.3 times 10 to the negative seven and 5.6 times 10 to the negative 11 respectively. So, win carbonic acid is involved. It is considered a dip protic acid, meaning that it can dissociate twice. And as such, if we set up our equation, we have the aqueous carbonic acid placed in a, placed in liquid water in a reversible reaction to dissociate and yield one aqueous proton an hour bicarbonate aqueous ion. Now, we can have yet another equation where we take our carbonate or bicarbonate in acres form place it in liquid water in a reversible reaction to yield another proton. And then we have our vinyl carbonate aqueous ion. Now, we have the acid ionization constants for the first step. And the second step. So if you were to set up the acid association constant expressions for our first step, that would be the ao ionization constant, which is equal to 4.3 times 10 to the negative seven is equal to the concentration of our products. So we have our proton, we multiply that by the concentration of the other product which is the bicarbonate. And then we divide that by the concentration of the reactant. So for the first step, that would be our car ban our carbonic acid. And then for our second reaction, we take the 2nd, 2nd acid cons the second acid ionization constant, which was equal to 5.6 times 10 to the negative 11. And we set that equal to the concentration of the products in the numerator which will be a proton. But this time we will have our carbonate. And then we would divide that by the concentration of our reactants which is our bicarbonate. So to convert the Ph into the concentration of hydro ions, we have to use the equation that the PH is equal to the anti log or negative log multiplied by the concentration of hydro ions. Now, we can rearrange this equation such that the concentration of hydro ions is essentially the anti blog negative log multiply by our negative at 5.8 that will come from the PH. And in doing so, we will arrive at 1.5 85 times 10 to the negative six molar. So we have the concentration of our hydro ions to set up the total car carbonate concentration equation. So the total amount which is t we will have the concentration of our carbonic acid plus the concentration of our bicarbonate. And then plus our concentration of our carbonate. And we will equal that to 1.05 or 1.0 times 10 to the negative fifth molar. Now, we need to have three separate equations to solve for the three unknown values or our first equation, we will take the first K A, we have our first K A and then we have our second one and then we have our third one. So to solve or to rewrite our expression for our first acid ionization constant in terms of bicarbonate, we have to rearrange it such that we go from the original to the following. And so now we are solving in terms of our carbonic acid, we would do the same thing for the second equation and we will write it in terms of our carbonate. So we will have our carbonate. Now, the concentration of our carbonate is equal to 5.6 times 10 to the negative 11 multiplied by the concentration of our bicarbonate and then divided by the concentration of our protons. Now, we can essentially summarize the new equations and solve for the concentration of our bicarbonate. First. So our concentration or our bicarbonate can be solved. It can be solved by taking 1.10 to the negative fifth, which we solved from a total carbonate concentration. And then we can divide that by 4.6 85. And so that 4.685 came from when we summarize the new equations and we solved in terms of bicarbonate. And this will give us a concentration of 2.134 times 10 to the negative six molar. So we have one of our concentrations, the all four hour carbonic acid and carbonate, we will have the concentration of our carbonic acid equal to. We had our equation set up earlier. We will have our 1.58 times 10 to the negative six multiply by 2.134 which is the concentration of our bicarbonate. This would be times 10 to the negative six. And then we divide that by 4.3 times 10 to the negative seven molar, which is the concentration that we found for or which is the acid ionization constant. And in doing this, we will have a 7.866 times 10 to the negative six molar. And then last but not least we have our, the concentration of our carbonate that we can solve for. So we could just use this equation that we've already drawn. And if we were to substitute, so we have our 5.6 times 10 to the negative 11, if you were to substitute the concentration of, of our bicarbonate, which we salved was 2.134 times 10 to the negative six. And then divide that by the concentration of our protons, this was 1.58 times 10 to the negative six. We would arrive at a concentration for carbon eight as 7.54 times 10 to the negative 11 molar. And so with that, we have solved the problem overall, I hope this helped. And until next time.

Verified Solution

Key Concepts

pH and Acid-Base Chemistry

Equilibrium and Ka Values

Experimental Methods for Analyzing Rain Composition