Rainwater is acidic because CO21g2 dissolves in the water, creating carbonic acid, H2CO3. If the rainwater is too acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, CaCO3). Calculate the concentrations of carbonic acid, bicarbonate ion 1HCO3-2 and carbonate ion 1CO32 - 2 that are in a raindrop that has a pH of 5.60, assuming that the sum of all three species in the raindrop is 1.0 * 10-5 M.

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Hello everyone. So in this video we're given some information about hydrogen sulfide, which is a weak byproduct acid. So we want to go ahead and calculate for the concentration of our hydrogen sulfide, the by sulfide ion and sulfide ion. And we are also given that the total concentration of these three species is 2.45 times 10 to negative five molars. So first I want to go ahead and write out my equilibrium dissociation um equations. So first we have our H. Two S going head to dissociate to the H plus, Canadians as well as the H. S. Minus and ions. The K. A. One value of this is 1.0 times 10 to the negative seven. Which is of course meaning that the P. K. A. Value is going to be 7.0. So we know that the K. One value is equal to products or the concentration of our products over the concentration of our reactant. So for the products we have our H plus multiplied by the H. S minus. And then from Iraq tents we have just one starting material which is the concentration of H. two s. So solving for the concentration of the age to us, we'll just do some manipulation to get us that the concentration of our H two S. Is equal to the concentration of H plus, multiplied by the concentration of H. S minus. All over K. A. One. The next equation that we have is going to be that the H. S minus will continue to associate into the H plus ions And the S. 2 -5. So I'll put the K. Value S. K. A. Two. That's going to be a value 1.3 times 10 to the negative 13 meaning that my P. K. A. Value for this is going to be 12.89. Same exact process. The K. two value is equal to the concentration of my products over my reactant. So for the product side we have our h. Plus and R. S. Two minus all over the concentration of my starting regents. In this case this will be a church s minus. Now I want to go ahead and isolate for the concentration of my S. Two minus. So with some mathematical manipulation We can get that. This will equal to the K. A. two multiplied by my H. S minus all over the concentration of my H plus. So scrolling down to give us a little bit more space we want to first calculate for the concentration of H. Plus we are given in the problem that my P. H. Is equal to 5.85. If you recall the concentration of my H. Plus, the equation for that is 10 to the negative ph which is just 10 to the negative 5.85. Putting that into my calculator, I will get the value of 1. times 10 to the negative six moller. Another part of the equation is that let's just scroll down first is that the total concentrations is equal to a certain value. So we have the three species H. To us H. S minus And the concentration of us to -2 equal to 2.45 times 10 to the -5 moller. So we solve for some of these values here which is H. Two S. And us two minus here. We'll just go ahead and put those equations into this formula. So we get for the first species that is the concentration of H plus multiplied by the H. S minus concentration over R. K. A. One value plus our concentration of H. S. Plus the formula for the S two minus which is K. A. Two times R H. S. All over the concentration of our H. Plus and that all equals to 2.45 times 10 to the negative five molar. So first step in solving for this is to multiply every term by our K. One value and then we're gonna go ahead and isolate the H. S minus concentration if we do so we get that the concentration of H. S minus will equal to 2.45 times 10 to the negative five, multiplied by R. K. A. One value multiplied by the concentration of H plus and this will be over our concentration of H. Two plus squared plus our K. One value multiplied. Bar H plus concentration. Go ahead and create that clearly. Okay and then we're gonna add our K. One value multiplied by R. K. A. To value. Now actually plugging in numerical values. Now we'll get that the concentration of H. S minus on the denominator. We have 1. times 10 to the negative six squared plus the K. One which is 1.0 times 10 to the negative seven. Multiplied by 1.41253 times to the negative six. And again we have our K. One value being 1.0 times 10 to the negative seven. Multiplied by R. K. Two value of 1.3 times 10 to the negative 13. And that's just the denominator on the numerator will rewrite this first term 2.45 times 10 to the negative five. Multiplied by my K. One value 1.0 times 10 to the negative seven. Multiplied by my concentration of my H. Plus candy on though that is 1.41253 times 10 to the negative six. So putting all this into my calculator I finally get the concentration of H. S minus two B. 1.62 times 10 to the negative six Mohler. Now we can go ahead and use this value to plug it back into my K equation. So we do so we get that the concentration of our H. Two S. Is equal to 1.41253 times 10 to the negative six. Multiplied by the concentration of H. S minus which we just all for is 1.62 times 10 to the negative six. And that's going to be over our K. one value. So 1.0 times 10 to the -7. Again putting this into my calculator, I'll get out the value of 2.29 times 10 to the - molar. So this is my concentration for my H. Two S. This is my concentration for H. S. Minus one more species. To go ahead and solve. Let's do this maybe on top here in green. So we're calculating finally for the concentration of R. S. Two minus an ion. So that will equal to 1.3 times 10 to negative 13, Multiplied by 1.6, 2 times 10 to the -6. In my denominator I have 1. times 10 to the negative six. So I'll just put this into my calculator and now I get the concentration value of 1.49 times 10 to the negative 13 bowler. And this is going to be my final answer for this problem.

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Key Concepts

Acid-Base Chemistry

Carbonic Acid Equilibrium

Concentration Calculations