![]() ![]() So our initial concentrations, our change, and finally ourĮquilibrium concentrations. Of silver chloride, let's go ahead and write our ICE table. So there are silver plus-oneĬations in solution, and there are also chlorideĪnions in solution. But some of it does dissolve, and so we get a saturated solution of silver chloride. So there's my little pile of undissolved silver chloride. So silver chloride is only a slightly soluble compound. ![]() And some of the silverĬhloride is going to dissolve, but most of it's going to So we take some solid silver chloride, so that's this over here, and we put it in some pure water. We're going to have some ammonia present. And in part B, it's no longer pure water. The molar solubility of silver chloride in pure water, and then we're going to compare the solubility that we get in part A to the solubility that we're going to get in part B. This would result in x = 0.00133, which is about 10x less than what we got for x when we approximated.Īnd if you analyze the math in this closely, you can conclude that the small x approximation is going to be invalidated when Ka is really large (i.e., a relatively strong weak acid) and the concentration is really small. To check if your approximation of 0.1-x = 0.1 is valid, divide x by 0.1 and see if it falls within 5%:īecause we exceed 5%, our approximation was invalid and we need to return and solve this problem using quadratic. Go ahead and approximate it to 0.1, then solve for x and you get x = 0.0134 Now that 0.1M-x on the denominator of the right hand side is the pain. Let's take the dissociation of 0.1M acetic acid (I'll call it HX for simplicity sake), a weak acid with Ka = 1.8x10^-5. It's used by comparing the x value you get, to the approximation you make in the first steps. A good guideline for the "small x approximation" is the 5% rule. ![]()
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