Predict whether a precipitate will form if you mix 75.0 mL of a NaOH solution with pOH = 2.58 with 125.0 mL of a 0.018 M MgCl₂ solution. Identify the precipitate, if any.

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Hello everyone. Today we are being asked to find the precipitate that is formed when 56 mL of K. O. H. R potassium hydroxide with the P. O. H. Of 1.67 is mixed with 138 mL of 0.16 Mohler barium chloride. So the first thing we have to do is we have to show the dissociation of both of these compounds. And so first we have K. O. H. We say that this dissociates into potassium plus ions as well as hydroxide ions. And then for barium chloride we say that be a cl two would dissociate into barium two plus ions noted by the B. A. Two plus as well as to chloride ions. And so the reason why it's two chloride ions is because when we have our starting material that has a subscript of two, we must have two in the resulting product. And so the next thing we have to do is we essentially have to cross these as if they were being bonded to one another. So that means we have to take barium and bonded with hydroxide and we have to take potassium and bonded with chloride. And so that's going to look a little like this. We're going to say barium two plus plus hydroxide is going to give us B. A. O. H. Or barium hydroxide to we have to cross the charges. So if we have a plus two charge herbarium that becomes a subscript for hydroxide and the minus charge minus one charge for hydroxide becomes a subscript for barium. And so barium hydroxide is actually going to be pretty insoluble in water. And so what this means is it's going to form a precipitate. So we're going to say it forms a precipitate for our potassium chloride. We say our K plus is going to be formed with our cl minus. And since both of their charges are one, it's just simply going to be K C. L. Which is actually very soluble in the solution, which is going to form an agreed solution as well. So we know we need to focus on our barium hydroxide. Next we have to set up our K. Sp expression. We're gonna have our K. S. P. Is equal to the concentration of barium two plus times our concentration of hydroxide minus. We're going to add a two for the exponent, since there's two as a subscript for hydroxide, that just becomes the exponent. And we're going to say that this is going to be equal to five times to the negative third. And so the K. S. P. Should be provided with you whenever you solve this on your own. And so we need to find the polarity of that barium two plus and similarity of hydroxide to plug this into our equation. And so for B. A. Or barium two plus, we're going to say barium two plus is equal to the polarity. Where we're going to say we have 0. moles of B A C. L. Two. And then we're going to put that over one leader. We're then going to multiply this by the multiple ratio between B a c L two and be a two. We're going to see that for every one mole of B A two plus, we have one mole of B a cl two. And so when we do this, our units are going to cancel out and we are then going to be left with, I'll draw it down here 0.16 polarity. And we're going to do the same thing for hydroxide. We're going to say oh h minus. And then we're gonna say since we already have the P O. H of that, we can go ahead and actually find clarity from that P O H. And by doing that we can say that the O. H of the hydroxide concentration is going to be 10 to the negative P O. H. We plug in that P. O. H. We're going to have negative 1.67. And so our concentration of hydroxide ions is going to be 0.2138 polarity. Now the final thing that we have to do is we have to find our q quotient Q is essentially the same thing as K. Except K is at equilibrium and Q. Is any point in the reaction. And so it's going to be the same Mechanism in which we do that. We're going to take our concentration of barium two plus which is 0.016 modularity. Multiply that by the hydroxide concentration which is 0.02138. And we're going to square that. And this is going to give us a value of 7.3134 times 10 to the negative six. And so we can looking at R K. Value R K. S. P. And looking at our Q value, we can say that Q is much less than R K. S. P. And if our Q is less than R K. S. P then no precipitate will form. So we're gonna say no precipitate will form no precipitate forms. And this is our final answer. I hope this helped. And until next time.

Predict whether a precipitate will form if you mix 75.0 mL of a NaOH solution with pOH = 2.58 with 125.0 mL of a 0.018 M MgCl₂ solution. Identify the precipitate, if any.

Verified Solution

Key Concepts

pOH and pH Relationship

Solubility Product Constant (Ksp)

Precipitation Reactions

Predict whether a precipitate will form if you mix 75.0 mL of a NaOH solution with pOH = 2.58 with 125.0 mL of a 0.018 M MgCl2 solution. Identify the precipitate, if any.

Verified Solution

Key Concepts

pOH and pH Relationship

Solubility Product Constant (Ksp)

Precipitation Reactions

Predict whether a precipitate will form if you mix 75.0 mL of a NaOH solution with pOH = 2.58 with 125.0 mL of a 0.018 M MgCl₂ solution. Identify the precipitate, if any.