Calculate the solubility (in grams per 1.00 * 102 mL of solution) of magnesium hydroxide in a solution buffered at pH = 10. How does this compare to the solubility of Mg(OH)2 in pure water?

Video transcript

Hello everyone today. We are being given the following question and asked to solve for it. So it says cobalt hydroxide is dissolved in the solution buffered at ph 9.5, calculate the scalability of cobalt hydroxide and grams per one times 10 to the second milliliters of solution. And then were asked to compare this to the ability of cobalt hydroxide pure water. So the first thing you wanna do is you want to draw cobalt hydroxide, ceo hydroxide being dissolved in water and when it does that it forms one cobalt two plus ions as well as to hydroxide aqueous ions. We then say the KS. P for this reaction is going to be 5.92 times to the negative 15th. Where they're going to set up our ice table which we can just do here, we could say our initial concentration for our cobalt hydroxide is going to be not applicable for our cobalt and a hydroxide, it's also going to be zero. But then we do plus X. For X. Being any given concentration and then we do plus two X for hydroxide. Since there's two of those ions. E. Is essentially I plus C. So we just have X. And then we have two X. Here setting R K sp up. We have K S p. Is equal to R five point Yeah R K S. P. That we're gonna be setting up for this is going to be our 5.92 times 10 to the negative 15th is equal to our concentration of cobalt two plus times are concentration of hydroxide and then since we have that coefficient of two, that's going to become our exponent in front of that. Behind that hydroxide simplifying this, we're going to get 5.92 times 10 to the negative 15th is equal to X times two X squared. This can be further simplified to 5.92 times 10 to the negative 15th is equal to four. X. Cute solving for X. We get that X is equal to 1.1396 times 10 to the negative five molar or our leaders. We can then take into account the molar mass for cobalt hydroxide. So the molar mass of cobalt hydroxide is going to be equal to 92.9492 g per mole. Setting up our equation to find the density or grams from milliliters, we can say that X is equal to one times 10 to the second power milliliters. We have to convert this into leaders first so we'll see that one middle leader is equal to 10 to the negative third Leaders. We want to multiply this by the X. That we found in the previous question which is 1.1396 times to the negative fifth moles per liter. Then we finally multiply that by the molar mass of 92.95 to shorten it out grams per mole. And so when our units cancel, we get an X value of 1.6 times 10 to the negative fourth grams per middle leader For the second half of this question, we know that our ph is equal to 9.5. And so to solve for the hydroxide concentration, we can simply raise the ph to a power of 10. So we do negative ph plugging in our PHP is negative 9.5. So our hydroxide concentration is going to be 3.16 times 10 to the negative 10 moller. R K constant. Kw constant is just saying that our concentration of hydroxide ions times our concentration of hi joanie um ions or H three plus ions is going to equal 10 or one times 10 to the negative 14th of course to rearrange this equation. To solve for our concentration of hydroxide ions, you can say that the concentration of hydroxide is equal to R K W. Over our H 30 plus ions take this equation over here, We're going to say that RKW is equal to one times 10 to the -14. And then we're going to divide this by our concentration of high ammonium ions which is 3.16 times 10 to the negative 10 moller. And that gives us a final concentration of 3. times 10 to the negative five for hydro hydroxide concentrations. We then have to set up a secondary ice table here, we see that our cobalt hydroxide over here can associate into once again cobalt two plus ions plus two hydroxide ions. We set up our ice table here. We see our initial concentrations for cobalt hydroxide don't matter for cobalt it's zero. But for hydroxide this time it's 3.16 to 3 times 10 to the negative five molar. Once again we just add plus X to the product side. And then for the hydroxide it's plus two X. So our equilibrium is going to be X for cobalt but it's going to be 3.16 to 3 times 10 to the negative five plus two X. For our hydroxide ions, we can assume that Negligible. So we can go ahead and get rid of that two X. We can then solve for R. K. S. P. On the right here which is 5.92 times 10 to the negative 15th is equal to our concentrations of cobalt and high sodium ions. So we're simply just going to say that's ex What you say, that's 3.16, 2, 3 times 10 to the -5. We're going to square that. We're eventually gonna end up with a value of X being equal To 5.92 times 10 to the -6 Mueller. Further solving this, we then have to take our value of X. Or the last step is to take that value of X. Which is equal to that one times 10 to the second milliliters convert that into leaders. So we're going to say that 10 to the negative third leader is equal to one middle leader. And we're gonna multiply that by 5.92 times 10 to the negative six moles per liter. And the last thing we do is gonna multiply that by the molar mass we're going to see the molar mass is equal to 92. g per mole. And this gives us a final value of 5.5 times 10 to the negative 5th g per middle liters. And so if we compare this With our initial one we can say that's eligibility is going to decrease. So solid ability is going to decrease here. And with that we have answered the question, I hope this helped. And until next time.