So your initial concentrations would be one molar for acetic acid, one molar for the acetate anion, and then pretty close to And one is the sodiumĪcetate that you add in to make your solution. So the acetate ions come from two sources. So really we have one molar, let's go ahead and put So we have another sourceįor acetate anions. But we have to add in something, right? Because we're also dealing with this, one molar concentration of sodium acetate. So we're going to gain a concentration of the acetate anion here. We lose for acetic acid, we would gain for the acetate anion, and therefore we would And then for our change, we said whatever concentration We started with our initial concentration, here we have a 1.00 molarĬoncentration of acetic acid. So these are the kinds of problems that we've been doing. And right now let's justįor a second pretend like we have only acetic acid. So we have acetic acid plus water, so we're going to haveĮverything at equilibrium, with our products H3O+Īnd acetate, CH3COO. So we're just gonna start by rewriting our acid-base reaction. So calculate the pH of a solution that is one molar in acetic acid. Let's go ahead and do the calculation and see that that is true. So we'd expect a pH that's higher than just a solution of acetic acid alone. The other source is the sodiumĪcetate that you added in. So one is the ionization of acetic acid, that's one source for your acetate anion. So there are two sourcesįor your acetate anion. So the acetate anion is the common ion, and this is the common ion effect. This decreases theĬoncentration of hydronium ion, and if you decrease theĬoncentration of hydronium ion, you're going to increase the Some of your hydronium ion when your equilibrium shifts to the left. Of this acetate anion is going to react with Going to shift to the left, and that means that some Of one of your products, the equilibrium shifts to the left. And according to Le Chatelier's principle, if you increase the concentration So you're increasing the concentration of one of your products. So if you add some sodiumĪcetate to your solution, you now have some more acetate anions. What would happen now if youĪdded some sodium acetate? Right, so if you added And so there's a concentration of acetate anions in solution. Acetic acid's going to donate a proton to H2O to form H3O+ or hydronium and the conjugate base toĪcetic acid, which is acetate. Question is, will you also obtain the correct pH value regardless which compound you use to react with water as long as you set up the reaction equation properly to either form H3O+ or -OH and use the Ka value for H3O+ and Kb value for -OH in order to solve for x and then further solve for pH or pOH? Or do you have to use the molecule that doesn't form a by-product CH3COOH vs CH3COONa, Na+ being considered the by-product, therefore CH3COOH favored likewise, NH3 favored over NH4+ with by-product NO3-? and solve for x, which = 1.3x10^-9 then solve for pH using x = 8.88 which is exactly what was obtained in the video. I then calculated Ka using (Ka)(Kb)=(Kw), with the given Kb value. Initial concentration = 0.35 nothing 0 0.15Ĭhange in concentration = -x nothing +x +xĮnd concentration = 0.35-x nothing x 0.15+x For the second example problem pertaining NH3 and NH4+NO3-, instead of having the NH3 react with water to form NH4+ and -OH, I had NH4+ react with water to form H3O+ and NH3.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |