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Ch.16 - Chemical Equilibrium

Chapter 16, Problem 51

Silver sulfate dissolves in water according to the reaction: Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq) Kc = 1.1 * 10-5 at 298K A 1.5-L solution contains 5.14 g of dissolved silver sulfate. If additional solid silver sulfate is added to the solution, will it dissolve?

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Hello everyone. So in this video we're going to go ahead and take a look at this equation right here and we're trying to see if we add additional sodium barium carbonate, which is our starting agent here to a certain polarity concentration solution. Would it still dissolve? So first let's go ahead and calculate firm polarity. So first we can start off by Kathleen for the molar mass of bearing carbonate. Baron carbonate is going to be B a c 03 and the molar mass that is going to be 1 97.34 g per mole. Now let's recall similarity Is equal to moles per leaders. The information that we're given from. The problem is that we have a 2.0 leader solution And then we have that 15.02 g of barium carbonate. So this right here will be volume and this right here will be mass. Alright? So since we know that this is what we're looking for and we're given this information right here, let's do some dimensional analysis then. So to fit this here instead of the moles will just have the grounds. So the mass for now. So that's going to be 15.02 g over two leaders. Now converting grounds to most, we can use the molar mass that we have calculated over here. So on the bottom we'll put the ground so it can cancel out and on top will be malls. And if you can see the crumbs cancel out perfectly. Putting this into my calculator, I will get a value of 0.038 polarity. Alright now scrolling down, We want to go ahead and calculate for our Q. All right, so our Q. Value is going to be the concentration of our barium Catalan multiplied by the carbonate, fly, atomic and ion. So in brackets this just means concentration. Alright, so we can say the concentration of our barium carbonate is actually equal to the concentration of its ion size. Composer. If that is true then then that all equals to 0.038 polarity. So now that we have this value and everything goes each other, let's go ahead and utilize this equation then. So that would be Q equals two hours. 0.38 polarity times our 0.38 polarity. So putting that into my calculator then I'll get a value of 1.444 times 10 to the negative three. So that's gonna be actually a lot greater than the value of R. K. C. Value. With this K. C. Is is an equilibrium constant. That's measured in moles per liter. If you look at the top we have our Casey value right over here. Alright, now that we know that our Q value is greater than our constant, then this means that the reaction shifts to the reverse direction to reach equilibrium. And what that means is that additional barium carbonates will not dissolve. Alright, and this is going to be my final answer for this problem. Thank you all so much for watching.
Related Practice
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Consider the reaction:

2 NO( g) + Br2( g)Δ2 NOBr( g) Kp = 28.4 at 298 K

In a reaction mixture at equilibrium, the partial pressure of NO is 125 torr and that of Br2 is 148 torr. What is the partial pressure of NOBr in this mixture?

Textbook Question

Consider the reaction: SO2Cl2(g) ⇌ SO2(g) + Cl2(g) Kp = 2.91*10^3 at 298 K In a reaction at equilibrium, the partial pressure of SO2 is 117 torr and that of Cl2 is 205 torr. What is the partial pressure of SO2Cl2 in this mixture?

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NH4HS(s)ΔNH3( g) + H2S( g)

At a certain temperature, Kc = 8.5 * 10 - 3. A reaction mixture at this temperature containing solid NH4HS has [NH3] = 0.0822 M and [H2S] = 0.0822M. Will more of the solid form, or will some of the existing solid decompose as equilibrium is reached?

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Consider the reaction and the associated equilibrium constant: aA(g) ⇌ bB(g) Kc = 4.0 Find the equilibrium concentrations of A and B for each value of a and b. Assume that the initial concentration of A in each case is 1.0 M and that no B is present at the beginning of the reaction. c. a=1;b=2

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For the reaction shown here, Kc = 255 at 1000 K. CO(g) + Cl2(g) ⇌ COCl2(g) If a reaction mixture initially contains a CO concentration of 0.1500 M and a Cl2 concentration of 0.175 M at 1000 K, what are the equilibrium concentrations of CO, Cl2, and COCl2 at 1000 K?

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