Consider a galvanic cell that utilizes the following half-reactions: (d) Will AgCl precipitate if 10.0 mL of 0.200 M NaCl is added to the solution in part (c)? Will AgBr precipitate if 10.0 mL of 0.200 M KBr is added to the resulting solution?

Question

Consider a galvanic cell that utilizes the following half-reactions:
 
<QUESTION REFERS TO APPENDIX B> 
(d) Will AgCl precipitate if 10.0 mL of 0.200 M NaCl is added to the solution in part (c)? Will AgBr precipitate if 10.0 mL of 0.200 M KBr is added to the resulting solution?

Accepted Answer

Hello. In this problem, we are told the following cathode reaction is used in a voltaic cell. The cathode solution was created by mixing 50 mL of 500.1 molar mercury to nitrate solution and 50 mL of a six mil ammonia solution. In part one, we are asked to determine if mercury to bromine precipitate will form upon the addition of 10 mL of a one times 10 to minus three Mueller sodium bromide solution. And in part two, we're asked to determine if mercury to ID I precipitate will form upon the addition of 10 mL of a one times miles three molar sodium iodide to the resulting solution from Part one were given the values for the formation constant of our complex sign. This liability product constants for mercury to bromide and that of mercury to I died. Beginning with Part one we want to find our initial concentration of mercury to and our initial concentration of ammonia and then our equilibrium concentrations of mercury too. Part one. Our total volume then Is 50 ml of mercury to nitrate plus 50 ml of ammonia plus ml of the sodium bromine. So this works out to a total volume of 110 ml. Our initial concentration of mercury to We take ml Of a .1 moller mercury to nitrate solution and our final volume is 110 ml. So our concentration is 0.04545 Mueller. Our initial concentration of ammonia. We take 50 ml. I'm a $6 dollar Ammonia solution and our final volume is 10 male leaders. So this works out to 2.7273 Moeller. We see that for both of these initial concentrations are units and milliliters cancels and we're left with our units of concentration. Now we'll set up a nice table related to the formation of our complex on. So we have mercury to reacting with ammonia for more complex iron. We have initial change and equilibrium. Initially we have 0.4545 moller mercury two and 2.7273 Mueller ammonia and initially none of our complex sign. Given the large value for our formation constant, we're gonna assume that 100% of the mercury reacts for every mole of mercury that reacts, we have four times that for the ammonia. And for every one mole of mercury that reacts, we form an equal amount of our complex iron. So at equilibrium we have no mercury two, we have 2.5455 for pneumonia. And we have 0.45454. A complex sign. We're going to assume a small back reaction. So then we get plus X plus four X. Based on the strike geometry and minus X. Once we reestablish equilibrium, we have X 2.5455 Plus four x. And the 0.04545 -6. Given a large value for our formation constant, we're going to assume that the X term is small relative to the remaining portion. So that would be our Equilibrium concentration before that small back reaction. So this is gonna be then approximately 2.5455 for the ammonia and approximately 0.04545 for the complex sign. Writing our formation constant expression we have in our equilibrium concentration of complex science. All over the concentration of mercury too, times of concentration of ammonia the fourth the scent is equal to 0.4545 divided by X times 2.5455 to the fourth power which we're told is equal to 1.8 times 10 to 9, Solving for X. We get X is equal to 0. 4545 divided by 1.8 times 10-9 Times 2.5455 to the 4th power. This works out to 6.014 times 10 to minus Mueller. This is our equilibrium concentration of mercury. To our next step then, is to find our reaction quotient and compare it to the scalability product constant. Sorry equation then that describes the equilibrium between are solid and its irons in solution. From this, we can write the scalability product costing expression equal to the concentration of mercury two times the concentration of squared Parasol does not appear in art product constant expression we're told this is equal to 6.2 Times 10 : -20. Our reaction quotient expression is written the same way we have a concentration of our mercury too intense concentration of squared. Finding the concentration of our bromide ions. We have then 10 ml. I'm the one Times 10 to -3 Mueller from iodine solution and Our final volume is ml. This works out to 9.09, 1 times 10 -5 Moller calculating our reaction quotient, we have the concentration of mercury ions which we found was 6.14 times 10 to minus 13. And now the concentration of our bromide ions that would be squared. So our reaction quotient q works out to 4. Times 10 -21, Which is less than 6.2 Comes 10s and 20 which is equal to our scalability product constant. Given that Q is less than our cellular product constant. That means then that no precipitate of mercury to bromide reform. Going on to solve part two, we do it in a similar manner to that for part one. So we're first gonna find our initial concentration of mercury to our initial concentration of ammonia and our equilibrium concentrations of mercury too, Part two, then our volume has changed again because we've added an additional amount of the Study my diet solution. It's our volume at this point now is 10 ml of the city. Might I solution plus original Harrington milliliters from part one. This works out 220 mL. Our initial concentration of mercury to We took 50 ml 0.100 moller Mercury to nitrate solution. And now our final volume of 120 ml. This works out to 0.04167 Moller. And our initial concentration of ammonia. We took 50 ml about six Mohler Ammonia solution and our foreign volume is 120 ml. This works out to 2.50 moller. We're now going to create a nice table again to describe the formation of our complex iron. We have mercury reacting with ammonia or more complex iron. We have initial change and equilibrium initially. Then the concentration of mercury two is 0.4167, that ammonia is 2.50. And initially we have none of the complex line. Again, given our large value for the scalability product constant. We're gonna assume that all the mercury reacts For every one mole of Mercury acts, we have four moles ammonia react. And for every mole of mercury that reacts, we produce an equal amount of our complex iron. So be no mercury left at equilibrium. We have 2.333 for the ammonia And we have 0.04167. Or a complex science, we're gonna again assume we have a small back reaction. This then will be plus X plus four X. Based on strike geometry and minus X. When we come back to equilibrium we'll have X 2. plus four X 0.4167 minus X. We're gonna assume. Then given a large value for a formation constant that the terms with X are gonna be small. And so at equilibrium our amount of Ammonia is approximately 2.333 and that of our complex iron and equilibrium is approximately 0.4167. Our formation constant then is given by the concentration of our product. So that of our complex iron. All over. The concentration of mercury too, concentration of ammonia The 4th power. This works out to 0.04167. All over X Times 2. to the 4th power and we're told that This is again equal to 1.8 Comes 10-9, Solving for X. We get X is equal to 0.04167 by by 1.8 times 10-9 times 2.333 to the four power. And this works out to 7.814 times 10 to minus 13 bowler. This is equal to our equilibrium concentration of mercury to As with Part one. Our 2nd step then it's going to be defined our reaction quotient and compare it to our scalability product constant. Next case then we now have the equation that describes the equilibrium between our solid mercury two, I died and its ions in solution. Our scalability product constant expression. Then we have the concentration of mercury too, times the concentration of iodide squared. And the pure solid does not appear in our scalability product expression. We're told that this is equal to 2.9 times 10 to minus 29. We write our reaction quotient the same way we have the concentration of the mercury to the concentration of squared. Finding the concentration of iodide ions. We have then 10 mL of a one times 10 -3 solar solution And our final volume is 120 ml. This works out to 8.333 times 10 to the -5 moller calculating our reaction quotient. Then we have the concentration of our mercury ions which we found a 7. times 10 to minus 13. And the concentration of our guidelines, which is 8.333 times 10 to minus five. And that's squared A reaction quotient. Q. then works out to 5.43 Times 10 to the -21 Which is greater than 2.9 Times 10 to the -29 which is our scalability product constant. So given that Q is greater than K. S. P. It means that precipitate of mercury two I died. Will form. Thanks for watching. Hope this helps

Consider a galvanic cell that utilizes the following half-reactions: (d) Will AgCl precipitate if 10.0 mL of 0.200 M NaCl is added to the solution in part (c)? Will AgBr precipitate if 10.0 mL of 0.200 M KBr is added to the resulting solution?

Verified Solution

Key Concepts

Precipitation Reactions

Solubility Product Constant (Ksp)

Dilution and Concentration Calculations