Calculate the molar solubility of calcium hydroxide in a solution buffered at each pH. a. pH = 4

Video transcript

Hello, everyone. Today, we have a following problem. Calculate the more solubility of calcium hydroxide and a solution buffered at each Ph. And so we have the first ph here being a Ph equal to four. So we can recall that the solubility product constant or KSP is the measure of solubility of solid ionic compounds and a solvent at equilibrium. So to find KSP, we must make note of our compound that we have, which is calcium hydroxide. So we have the following formula for that compound and it would dissolve into calcium two plus aqueous ions as well as two hydroxide. And the reason why it associated in two hydroxide aqueous ions is because of that subscript two, letting us know that there are two of those hydroxides. So our KSP will be the concentration of our calcium ions multiplied by the concentration of our hydroxide ions. Now, it's important to note that that the amount of hydroxy ions becomes the exponent of the substrate that is attached to. So we have an X amount of two for the hydroxide. No, we can also recall that our equilibrium constant for these reactions is equal to the concentration of hydro ions multiplied by the concentration of hydroxide ions. And of course, our equilibrium constant here is one times 10 to the 14. So how do we find the concentration that we need? We also want to make note that to find our concentration of hydro ions to find their concentration of hydro ions, we can find this by taking BP and equaling it to the log of our concentration of hydro ions which also equal to 10 race of the power of negative Ph. So if we take 10 eras of the power of negative Ph, that's negative four, that gives us one times 10 to the negative four times 10 to the negative fourth molar. So if we plug that into our equation, and if we solve for a concentration of hydro ions, we get a concentration that is equal to one times 10 to the negative 10 molars. So the next step is to set up what's known as an ice table. So we have our calcium hydroxide that associates into our calcium ions as well as our two hydroxide ions. So we have our initial concentrations of each substrate. Of course, it will be zero for our solids as solids do not participate in these reactions. And then we have our calcium, which is what we need to find. So we label it as X and then we have our hydrox sign which is one times 10 to the negative 10 molar. Our change will simply be subtracting X from our reactants, such that we can add X to both of our products. Then more specifically, this will be two X because we have two of hydroxide ions and then we have the final E where we have our X. So now that we have set up this table, we can now, so for the molar solubility, so we had our KSP which we had was our 4.68 times 10 to the negative six. We equal that to our concentration of our S which will be substituted as our X, it'll be multiplied by the concentration of one times 10 to the negative 10 raised to the X one F two. So for a concentration of S we get 4.68 times 10 to the negative 14th molar. And if we round that to the nearest number, we get answer choice a five times 10 to the 14th molar is our answer overall. I hope this helped. And until next time.