(b) Will Mg(OH) 2 precipitate when 4.0 g of Na 2 CO 3 is added to 1.00 L of a solution containing 125 ppm of Mg 2+ ?

Hi everyone for this question we're being asked if 3.5 g of potassium carbonate is added to 1.50 liters of a solution containing 156 parts per million of ferrous ion will iron to hydroxide precipitate form. Okay, so our goal here is to determine whether or not precipitate will form and that specific precipitate is iron to hydroxide. So we need to first start off by writing out a reaction. Okay, so we know we have iron to hydroxide and when this breaks down it is the ferris ion Plus two hydroxide ions. So this is our balanced chemical equation. And what we're going to want to do now is find out our values for our equilibrium constant and our reaction quotient and compare those two to see if precipitate will form. So our equilibrium constant. K represents our when concentrations are at equilibrium, our reaction quotient Q is non equilibrium concentrations. So when K is greater than Q, we say the reaction proceeds in the forward direction. Okay, And I normally like to write K first because it creates this arrow here and that lets you know which direction it's going to proceed. So when K is greater than Q, it proceeds in the four direction, when K is less than Q, the reverse direction is favored and when K is equal to Q, that means we're at equilibrium. So we want to know out of these three options which one will cause iron to hydroxide precipitate to form. If we look at our balanced equation, we see that the iron to hydroxide is a reactant and so in order for precipitate to form, we're going to want to create more reactant and that will be favored by a reverse direction. So when we reverse the direction that means we're going to shift to the left and so by shifting to the left, our concentration of products is going to decrease and our concentration of reactant is going to increase. So now we know that we want K. To be less than Q. And so we can look up the value for K. And the value for K. For this reaction is going to be K. Is equal to 4. Times 10 to the -17. And this is for our equilibrium concentration of our product. Okay so remember when we write out our equilibrium expressions, its concentration of products over concentration of reactant. But because our iron to hydroxide is a solid because it's going to be a precipitate, we don't include that in our equilibrium expression. So R. K. Is going to equal 4.87 times 10 to the negative 17. And remember our Q. Is going to equal that's what we're looking for but it's going to equal the same thing. But this is non equilibrium concentrations. So now that we know our K. Value, we just need to figure out what Q. Is and then compare the two. Okay so the way that we can calculate Q. Is by first determining the concentration of ferrous ion. Okay so we have our fairest ion and we want the concentration of that. And remember concentration is more over leader or malaria. T okay so let's start off with what we were given in the problem when it comes to this ferris ion we were told we have 100 and 56 parts per million of the ferris ion. And what that means is we have 100 and 56 mg of ferris ion per kilogram of solution. So our goal here is to go from milligrams. Affairs ion per kilogram of solution to moles over leader. So let's go ahead and get started. Okay so to go from milligrams to grams our conversion is and one g Affairs ion There is 1000 mg of ferrous ion. Okay so making sure our units cancel here are milligrams cancel. And we're left with grams but we want to be in moles. So we need to go from grams to moles using molar mass in one mole affairs ion. Looking at the periodic table. The molar mass is 55 point 845 g of ferrous ion. So now our grams cancel. And we're left with moles in the numerator which is what we want. The last thing we want is our volume and leaders Okay so we have one kg of solution and we need to go to leaders of solution in one kg of solution There is one leader of solution. So you can see here that our kilograms of solution cancel and we're left with leaders of solution and we're in the correct units because we canceled out properly. So we are in moles over leader and that is concentration. So once we do that calculation we get 2.793 times 10 to the negative three moller of ferris ion. Okay, So now that we know our concentration affairs and we need to determine our concentration of hydroxide. Okay, So let's go ahead and do that as step two. So now we want to know our concentration of hydroxide ions. Okay, and this is going to be the same thing in moles over leader or polarity. So in the problem we're told we have potassium carbonate. So that's where we're going to start here and I'll tell you why in a minute. So let's switch up the color. So we're going to start with this potassium carbonate because once it breaks down it breaks down into potassium ions and carbonate ion. And this carbonate ion is a strong bronze said Lowry base which produces hydroxide ions in aqueous solution. So if we take this carbonate ion and we have it undergo hydraulic sis so it's going to react with water. It is going to produce bicarbonate and hydroxide. And we're interested in this hydroxide here finding out the concentration. So what we can do is we can look up the KB value for our strong Bronston. Lowry base and the KB value for it is K B is equal to 1.8 times 10 to the negative four. So what we can do here Is we'll start off with our 3.5 g of potassium carbonate and find out our concentration of carbonate from that. Okay, so let's go ahead and do that. We're told we have 3. g of potassium carbonate per one liter of solution. So remember, our goal is to go from grams of potassium carbonate per liters of solution. Two moles of carbonate over Volume are leaders of solution. Okay and one mole of potassium carbonate. Our Molar mass is 138.205 g of potassium carbonate. So now our g cancel. And we're in moles of potassium carbonate. But we want to be in molds of carbonate. So we need to do a multiple ratio of potassium carbonate to carbonate. Okay, so in one mole of potassium carbonate we have one mole of carbonate. Okay, so our moles of potassium carbonate cancel. And we're left with moles Of carbonate over liter of solution, which is what we want. Okay, so when we do that we get 0. moller carbonate. Okay, so we're in the last stretch here. Now what we can do is using the KB value we were given we can write out the KB expression and solve for X, which will give us our hydroxide concentration. Okay, so we set our KB Is equal to 1.8 times 10 to the -4. And what we're doing is we're going to write this out as our concentration of products over concentration of reactant is raised to their exponents. So for our potassium carbonate reaction, our products are bicarbonate and hydroxide and our reactant is carbonate. Okay? And this translates to X. So our bicarbonate and hydroxide are both X. So we'll have X squared in the numerator and the denominator. We just solved for our concentration of carbonate. So we'll plug that in. So we have 0.2532. So we have 1.8 times 10 to the negative four is equal to X squared over 0.2532. So now we're just solving for X. And then we can plug into Q and figure out what our value is for Q. So let's square root both sides. And we'll multiply both sides of our equation by 0.2532 and then take the square root. And when we do that we get X is equal to 2.1351 times 10 to the negative three. So now we know what X. Is and we can calculate Q. Okay, so up above we said Q. Is equal to question mark. Are Faris ion concentration times are hydroxide concentration. So our Faris ion concentration. We calculated it to be 2.7934 times 10 to the negative three. And our hydroxide concentration we said is 2.135 times To the negative 3rd. And this is going to be squared because remember we said we have to raise it to its exponent and because we have two moles of hydroxide, we need to raise that to the second power. Okay, so we're going to get Q is equal to 1.27 times 10 to the negative eight. So now we know our Q and we know our K. So we need to compare the two values. Okay, so our K is 4.87 times 10 to the negative and r Q is 1.2734 times 10 to the negative eight. So in this situation K is less than Q, which means the reverse direction is favored and precipitate will form. So the answer to this problem is yes, because K is less than Q. The reverse direction is favored. So it's going to shift towards the left and precipitate will form. Okay, so we'll go ahead and highlight this here as our answer and this is the explanation to that. That's the end of this problem. I hope this was helpful

Ch.18 - Chemistry of the Environment

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Brown 15th Edition

Ch.18 - Chemistry of the Environment

Problem 70b

Chapter 18, Problem 70b

(b) Will Mg(OH)₂ precipitate when 4.0 g of Na₂CO₃ is added to 1.00 L of a solution containing 125 ppm of Mg²⁺?

Verified step by step guidance

Convert the concentration of Mg^{2+} from ppm to molarity. Since 125 ppm is equivalent to 125 mg/L, convert this to moles by dividing by the molar mass of Mg (24.31 g/mol).

Calculate the moles of Na2CO3 added by converting 4.0 g of Na2CO3 to moles using its molar mass (105.99 g/mol).

Determine the concentration of CO3^{2-} ions in the solution after adding Na2CO3. Since the volume of the solution is 1.00 L, the concentration is equal to the moles of Na2CO3.

Use the solubility product constant (K_{sp}) for Mg(OH)2 to determine the ion product (Q) of Mg^{2+} and OH^{-} in the solution. The K_{sp} expression for Mg(OH)2 is K_{sp} = [Mg^{2+}][OH^{-}]^2.

Compare the ion product (Q) to the K_{sp} value for Mg(OH)2. If Q > K_{sp}, Mg(OH)2 will precipitate; if Q < K_{sp}, it will not.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is a numerical value that represents the equilibrium between a solid and its ions in a saturated solution. For Mg(OH)2, Ksp indicates the maximum concentration of Mg2+ and OH- ions that can exist in solution before precipitation occurs. Understanding Ksp is essential to determine if the concentration of ions exceeds this value, leading to precipitation.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react to form an insoluble compound, or precipitate. In this case, adding Na2CO3 introduces carbonate ions, which can react with Mg2+ ions to form Mg(OH)2. Recognizing the conditions under which precipitation occurs is crucial for predicting whether a solid will form in a solution.

Concentration Calculations

Concentration calculations involve determining the molarity of ions in a solution, which is essential for assessing whether a precipitate will form. In this scenario, converting the mass of Na2CO3 to moles and then calculating the resulting concentration of Mg2+ ions allows for comparison with Ksp. Accurate concentration calculations are vital for predicting the outcome of the reaction.