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Course: CHEM 114, Fall 2008

School: Ill. Chicago

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Chemistry 112 Worksheet goal: Worksheet - Solubility Equilibria Harwood Applying your understanding of equilibrium reactions to solubility reactions. Work together in your groups. Feel free to use your TA or text as a guide, if needed. One of the reaction types that we discussed in Chem 112 was dissolution. Remember: CuBr(s) Cu + (aq) + Br (aq) Just like with many of the reactions we have been discussing,...

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112 Chemistry Worksheet goal: Worksheet - Solubility Equilibria Harwood Applying your understanding of equilibrium reactions to solubility reactions. Work together in your groups. Feel free to use your TA or text as a guide, if needed. One of the reaction types that we discussed in Chem 112 was dissolution. Remember: CuBr(s) Cu + (aq) + Br (aq) Just like with many of the reactions we have been discussing, some dissolution reactions don't go to completion (all the solid does not dissolve just some of it). This occurs for compounds that are only slightly soluble in water. These dissolution reactions exist in an equilibrium state. Some of the solid dissolves and some remains in the solid form. So we can rewrite the above dissolution reaction as: CuBr(s) Cu + (aq) + Br (aq) Just like any other equilibrium reaction, there is an equilibrium constant (K) associated with it. In this case we give the equilibrium constant a special symbol to distinguish it from other equilibrium constants. We call it a Ksp (for solubility product). These Ksp's are just the same as regular K's. We write the equilibrium expressions just like we would for any other equilibrium: Ksp = [Cu+][Br] (Remember that the solid CuBr is not included because it is SOLID.) They have the same properties as the K's we have studied already. A large Ksp suggests lots of products and a small Ksp suggests fewer products. So a compound with a large Ksp would be more soluble than one with a small Ksp (since the products of this reaction is the DISSOLVED CuBr and the reactant is the UNDISSOLVED CuBr). If we think about LeChatelier's Principle then we can think about things that will affect the solubility of a particular compound. So for the CuBr equilibrium above, if we add to that solution some Br then we would be adding a product and we would expect the equilibrium to shift to the left. Since the left side of the equation is solid CuBr then adding Br would make more solid CuBr and use up some dissolved CuBr. Therefore Br adding would decrease the solubility of CuBr. We could add Br to the solution by adding some NaBr, for instance. This is formally termed the common-ion effect because NaBr has an ion in common with our slightly soluble salt CuBr. That ion is obviously the bromide ion. Take this equilibrium: Mg(OH)2(s) Mg 2+ (aq) + 2 OH(aq) OH is one of the products that results from the dissolution of Mg(OH) 2. If we add some strong base to this + solution (such as NaOH which breaks up into Na and O OH ions in water) then we are adding extra OH to the solution. OH is one of our products. LeChatelier's Principle says that the reaction will shift to the left toward the INSOLUBLE reactant. Adding a strong base (or increasing the pH) to the solution of Mg(OH) 2 will decrease the solubility of Mg(OH) 2. This is an example of the effect of pH on solubility. Think about what would happen if you add acid (reduce the pH) to the magnesium hydroxide mixture. Acid will react with the base Mg(OH)2 to form a salt plus water. This will take away some of the OH from the solution. That is taking away some product so the reaction will shift to the right to make more SOLUBLE products and use up more INSOLUBLE reactants. Adding an acid will increase the solubility of the Mg(OH)2. 1. Write a balanced chemical equation for the equilibrium occurring when each of the following are added to water, then write the Ksp expression. a. Lead(II) carbonate b. Nickel(II) hydroxide c. Strontium phosphate Worksheet 9 7 Harwood 2. Which is more soluble in water at 25C, CuBr (Ksp = 5.3 10 ) or CuCl (Ksp = 1.9 10 )? Explain why. 3. Would you expect more CaCO3 (Ksp = 3.8 10 ) to dissolve in pure water or in a 0.01 M CaCl2 solution? Explain your answer. 9 4. Write a general solubility product constant expression for a salt of the form Ax By . 5. Calculate the molar solubility of silver thiocyanate (AgSCN; Ksp = 1.0 10 water containing 0.01 M NaSCN. (Use an ICE table.) 12 ) in pure water and in 2

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Ill. Chicago - CHEM - 114
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help simpson returns an approximation for the integral of f(x) over [a,b] using Simpson's rulesimpson('cos',pi/4,pi/2)ans = 0.5859diary off;
Ill. Chicago - MCS - 320
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Ill. Chicago - MCS - 320
% 1. polynomials: coefficient vectorsc = [ 3 -2.23 -5.1 9.8]c = 3.0000 -2.2300 -5.1000 9.8000y = polyval(c,0)y = 9.8000x = -1:0.01:+1;y = polyval(c,x);size(y)ans = 1 201plot(x,y)% 2. Polynomials defined by roo
Ill. Chicago - MCS - 320
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Ill. Chicago - MCS - 320
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Ill. Chicago - MCS - 320
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Ill. Chicago - MCS - 320
% 1. sparse matricesa = sparse(4,6,3,10,10)a = (4,6) 3spy(a)% a more interesting plot is generated by% the so-called Pascal matrices:p32 = pascal(32);s32 = rem(p32,2); % creates a sparse 0/1 matrixspy(s32)% to reduce this matrix
Ill. Chicago - MCS - 320
c = [3 -2.23 -5.1 9.8]y = polyval(c,x)x = 1y = polyval(c,x)x = -1:0.1:1;y = polyval(c,x);plot(x,y)r = roots(c)cr = poly(r)norm(c(1)*cr-c)plot(r,'bx')dc = polyder(c)hold onplot(roots(dc),'rx')yxc = polyfit(x,y,3)c3 = [0.74 0.97 1.1 0.
Ill. Chicago - MCS - 320
% we start with polar plots, plotting the spiral r = 0.01*t^2t = 0:0.1:10*pi;r = 0.01*t.^2 - 0.02;polar(t,r)% three dimensional plots require a grid for a surfacexa = -2:0.2:2; % division of the first coordinate axisya = -2:0.2:2; % division of
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