What is the solubility of Cr(OH)₃ at a pH of 11.10?...

Question

What is the solubility of Cr(OH)₃ at a pH of 11.10? (K^sp Cr(OH)₃ is 6.70 × 10^-31)

Accepted Answer

hi everyone for this problem it reads what is the soluble itty and grams per liter of nickel to hydroxide with a solid ability product of six times 10 to the negative 16 in a solution buffered at ph equals 5.85. Okay, so our goal here is to find the sala bility in grams per leader. All right. And so let's go ahead and start off by writing our equilibrium that is established between are solid and the ions and solution. Okay, So we have our nickel to hydroxide and this is our solid. And at equilibrium it produces nickel ions and hydroxide ions. And were given the value of the equilibrium constant K. S. P. And Casspi represents the product of the molar concentrations of our ions. Okay, so this product is going to be our nickel ions times are hydroxide ions and we're paying attention to our coefficients here because we have two hydroxide, we're going to square this here And in the problem we are told what r K. S. P. Value is and that value is 6. times 10 to the negative 16. Okay. Our goal here is to calculate the sala ability and grams per leader. Okay, and so here with our K. S. P value in order for us to solve for solid ability, we're going to want to figure out what is our concentration of hydroxide. Okay, and were able to calculate our concentration of hydroxide from one key thing that's given in the problem. Okay, so if we can remember our concentration of hydroxide is equal to 10 to the negative p. O. H. Okay, so remember we have a ph scale and we have a P O. H scale. And so our goal here is to solve for the concentration of hydroxide. And one thing to remember is that P O H is equal to 14 minus P. H. All right. And we know what our ph is because we're told in the problem. So our P O. H. Is going to be 14 minus 5.85, which gives us a P. O. H. Of 8.15. So let's go ahead and plug that in. So we get 10 to the negative P O. H. We said our P O H is 8.15. Okay, so that gives us a hydroxide concentration of 7.795. Oops, put a lot of space there. So 7. times 10 to the -9 moller. So this is our hydroxide concentration. So let's go ahead and solve for our nickel concentration. Okay, so our concentration of nickel ions we're gonna say is equal to X. Okay, and from here we're going to solve for X. Okay, so we know that we have our K sp expression is six point oh times 10. So the negative 16 is equal to X. Which is our concentration of nickel times our concentration of hydroxide which we just solved for. So 7. Times 10 to the negative nine squared. Okay, so this is basically this rewritten. Alright, but we're substituting s for a concentration of nickel. Alright, so our goal here is to solve now for X. Alright. So we'll first isolate it by dividing or isolating X by dividing both sides by 7.0795 times 10 to the -9 squared. All right. And when we do that, we get 11.971 molar. So let's go ahead and write that here are concentration of nickel which we let X be its representative is equal to 11.971 molar. Alright, so now that we know our concentration of hydroxide and we know our concentration of nickel. We can calculate the solid ability. Alright. And we can calculate the solid ability because our concentration of nickel ion is equal is the same thing as our concentration of our nickel to hydroxide. Okay, so by knowing what our concentration of nickel is, we'll be able to figure out our soluble itty. So let's go ahead and we know that our concentration of nickel ions is 11.971 Moeller. Remember molar is moles over leader and it's specifically moles of nickel. Two hydroxide. Okay, so we want to go from moles per leader. two g per leader. Okay, this is our goal. So let's go ahead and do that. Alright, so and one more of nickel to hydroxide. We need the molar mass. The molar mass is 92.71 g. So making sure our units cancel here are moles cancel and we're left in grams per leader, which is what we're looking for. All right. So once we calculate this out, we're going to get a final answer of 1.1 times to the third grams per leader. Alright, This is our final answer. This is the solid ability in grams per leader. That's it for this problem. I hope this was helpful.

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What is the solubility of Cr(OH)₃ at a pH of 11.10? (K^sp Cr(OH)₃ is 6.70 × 10^-31)