The solubility of CaCO3 is pH dependent. (b) Use the Kb expression for the CO32 - ion to determine the equilibrium constant for the reaction CaCO31s2 + H2O1l2 ΔCa2 + 1aq2 + HCO3-1aq2 + OH-1aq2

Video transcript

hi everyone for this problem it reads if the soluble itty of strontium chrome eight is dependent on ph what is the equilibrium constant for the following reaction based on the KB expression of the chrome eight ion and were given the K. A. Value the asset, which is the acid dissociation constant along with the equilibrium reaction. Okay, so our goal here is to solve for the equilibrium constant and this equilibrium constant is K. So we want to solve for K and K is going to be equal to our solid ability product or K. S. P times K B. Okay, so we need both K. S. P and K B to solve for this. So let's go ahead and get started. So the first thing that we're going to want to do is find out what our strontium chrome eight looks like in its equilibrium. Okay, so we have our strontium strontium chrome eight which is a solid. And at equilibrium with its solution we have the strontium ion And the Chrome eight Ion. Okay. And R K S P or are scalability product is going to equal the product of the molar concentrations of the ions And this is a value. We're going to look up in this problem. Look up for this problem since it was not given and this value is 3.6 times 10 to the -5. So now we see we have our scalability product. The next thing we need is our KB or are based association constant and an equation that relates K A. And K B is K W is equal to K A times K B. So our goal here is to solve for K B. So we'll go ahead and divide both sides by K A. And we get K B. Is equal to K W over K A. Okay, so kw is a constant. We should have memorized and that is one times 10 to the negative 14. We were given K A. In the problem which is our acid dissociation constant. And that value is three times 10 to the negative eight. So when we solve for K B we get 3. times 10 to the negative seven. Okay, so now we have both are solid ability product R K S. P and R K B. So let's go ahead and plug it in here. So we can solve for our equilibrium constant. K. Alright, so our K is equal to R K S P. Which we said is 3.6 times 10 to the negative five. And this was the value we looked up and our KB which we solved four is 3.33 times 10 to the negative seven. Okay, so we're gonna go ahead and solve and do this math. And when we do we get 1.2 times 10 to the negative 11 is our equilibrium constant for this problem. And that is the end of this problem. I hope this was helpful