A solution containing sodium fluoride is mixed with one containing calcium nitrate to form a solution that is 0.015 M in NaF and 0.010 M in Ca(NO3)2. Does a precipitate form in the mixed solution? If so, identify the precipitate.

Video transcript

Hello everyone today we are being asked to find the precipitate that is formed when .012 molar of sodium sulfate is mixed with .018 Mueller of barium nitrate. So the first thing you have to do is dissolve both of these or draw them out as if they were being dissolved. So we have any two S. 04. We're gonna say that dissociates into sodium ions and sulfate ions. Next we're gonna do this thing with barium nitrate. So we're gonna say B. A. N. 03 to dissociate into barium ions as well as nitrate ions in 03 minus. And so we can say that sodium plus an +03 is soluble together And that be berry. Um two plus and sulfate. And so for two minus form barium sulfate. Which is going to give us our precipitate. That is what is going to form our solid. So precipitate. We then therefore must determine the K. S. P. For barium sulfate. And so the K. S. P. For barium sulfate is going to be 1.5 times to the negative ninth. When we set up our Q. Quotient we're gonna say Q. Is equal to bury them. So be a two plus, multiplied by S. 042 minus. All right. And so we have to determine the concentration of barium two plus. We must take the number of moles that we have from our polarity which is going to be 0.018 moles of that. And then we're going to multiply by the multiple ratio. So we're going to say that for every one mole of barium nitrate, We have one mole of barium two plus. And this is gonna give us 0.018 moles Per liter of barium two plus. And then as far as S. 02, is that so two minus? We're gonna do the same thing. We're gonna say 0.12 moles. This is going to be for sodium sulfate. We're going to say for every one mole of sodium sulfate, We have one mole of just sulfate itself. So this is going to give us 0.01, two moles over liters of sulfate plugging us into our Q. Quotient. We have a Q. Is equal to 0.018. We're gonna multiply that by 0.12 to give us 2.16 times 10 to the negative fourth. And so since we have our Q. Here, we have our K. S. P. Here we can note that our Q. Is greater than R. K. And therefore a precipitate. So precipitate will indeed form. And this is our final answer. I hope this helped ahead Until next time