The precipitation of Al(OH)₃ (K_sp) = 1.3⨉10^-33) is sometimes used ot purify water. (a) Estimate the pH at which precipitation of Al(OH)₃ will begin if 5.0 lb of Al₂(SO₄)₃ is added to 2000 gal of water

Video transcript

Hello everyone. So in this video we need to find the correct ph that is needed to precipitate R. N. I. O. H. Two. And were given the K. S. P. Of N. I. O. H. Two. So first we can go ahead I route the equation. So we have their N. I. Oh H. Two but go ahead and associate into N. I. Two plus and two moles of our age minus. Of course we are given the K. S. P. value and that is a 5.48 times 10 to the -16. And the formula for the KsB of this reaction is that the concentration of N. I. Two plus is being multiplied with the concentration of R. O. H minus squared. So let's go ahead and do some dimensional analysis now that we have the reaction and the relating K. S. P. So We start off with the calculations that we have are given £3.5. Let's write this little bit more clearly 3.5. So £3.5 of R. N. I. And 32. Which gives. And then we have for our volume of our salute. Our solution to be gallons of water. Now we're gonna go ahead and change our pounds two g. The direct um conversion for this is that for every pound we have 453.6 g. Then we'll go ahead and use the molar mass of R N. O. N. 03 to to go ahead and give us of course are moles. So we can say that for every one mole of R N I N 032 we get 82.71 g of R N I N O 3 2. Let's just cancel out our units. We don't get lost. So we have the pounds and then we have our grams counseling out. Okay, let's continue forward. We can use the multiple ratio. Now we can based off the multi mole ratio. Using this equation actually is going to be the basically same. Just that R O H is different. But for our reactant and the molds of the N I two plus that's exactly. So we have one mole of R N. I two plus with one more of RNIN 032. So again canceling our units, the molds will cancel. So now we have the moles of N. I two plus. So I'm gonna go ahead and now convert my volume, my units of volume to go to leaders. So first were given or starting off with our gallons, I would move that into courts. So for everyone gallons with four quarts and for everyone court That is 0.946 L. You see that the gallons will cancel out and our courts will cancel out. So now putting all these numerical values into my calculator, I'll get the numerical value of 4.5926, 10, 10 to the negative four Moeller of N I two plus. And we have this concentration unit because we have leaders and molds mold over leaders will give us these capital M. Unit. So now the precipitation only occurs when our Q value equals to R. K. S. P. Value. So then for our cue that equals to let's see our ksB values given to us to be 5.48 times 10 to the negative 16. And that is equal to let's see we calculate for the concentration of N. I. Two plus earlier in green that is 4.5926 times 10 to the negative 11. Then for O. H. Concentration we still don't know this and that is going to be of course squared. So now we're just simply isolating the concentration of R. O. H minus if we do. So let's see here we have the concentration of R. O. H minus equaling. Then on a numerator we have 5. times 10 to the negative 16. That's over the concentration of R. N. I. Two plus which is 4. times 10 to the negative 11. And because we have a squared we'll go ahead and square root it then to undo that Of course I'll put this into my calculator I'll get the value of 2.0035 times To the -12. So if we want to go ahead and now use that to find our p. o. h. We know that the equation for P. O. H. Is that the negative, we took the negative log of the concentration of O. H minus. So putting that into my calculator, I'm just basically substituting this value into this. So put that into my calculator. I get the p. O. H. Of then 5.96. And now finally to solve for our ph how we get that is 14 minus the P. O. H. Value, which is 5.96. We will get that our P. H. Value. Then it's going to be eight point oh four. So this is my final answer for this problem.