according tp the reaction Cu^2 + 2NH3 ===> Cu(NH3)2^2+ 6.00 ml of 0.00500 M Cu(N
ID: 731079 • Letter: A
Question
according tp the reaction Cu^2 + 2NH3 ===> Cu(NH3)2^2+6.00 ml of 0.00500 M Cu(NO3)2 were diluted w/ water to final volume of 250.0 ml. Then, 5.00 ml of the diluted Cu(NO3)2 were mixed w/ 5.00 ml of NH4Cl and 5.00 ml of H2O.
--calculate the "initial" concentration of copper ions in the mixture.
can anyone help, i know we supposed to use M1V1=M2V2 but dont know which one to use
Explanation / Answer
1. calculate the moles of species: Mole Cu(NO3)2 initial = 2.00 L x 2.00 M = 4 mol Mole KOH initial = 2.00 L x 3.00 M = 6 mol 2. Write the reaction: Ionitation of Cu(NO3)2 Cu(NO3)2 -> Cu2+ + 2 NO3+ 4 mol 4 mol 8 mol Ionitation of KOH KOH -> K+ + OH- 6 mol 6 mol 6 mol The complete reaction: Cu2+ + 2(NO3-) + 2K + 2OH- -> Cu(OH)2 + 2K+ + 2NO3 Ion NO3- and K+ can be eliminated so the final reaction is: Cu2+ + 2OH- -> Cu(OH)2 The precipitate is Cu(OH)2 and the ions that remind in the solution is K+ and NO3-. Now we can calculate the mole of Cu(OH)2 that formed: reaction Cu2+ + 2OH- -> Cu(OH)2 Initial 4 mol 6 mol 0 mol Reacted 3 mol 6 mol 3 mol Equilibrium 1 mol 0 3 mol Note: We use moles of OH- for the limitiation, because if we use moles of Cu2+, they would be not enough because we need 8 mol of OH- since just provided 6 mol of OH- ( from 2/1 x 4 mol = 8 mol) 4. calculate the mass of precipitate. Moles Cu(OH)2 is 3 mol and its molar mass is 97.566 Mass Moles Cu(OH)2 = mole x molar mass = 3 x 97.566 = 292.698 g 5. calculate the concentration of ions left in the solution. Remember the ions that left in the solution is NO3- and K+ and Cu2+ that excess from the ionitation reaction of Cu(NO3) we get the mole of NO3- is 8 mol and from the ionitation of KOH we get the mol of K+ is 6 mol, and Cu2+ is 1 mol Cu(NO3)2 -> Cu2+ + 2 NO3- 4 mol 4 mol 8 mol KOH -> K+ + OH- 6 mol 6 mol 6 mol Thus: [NO3-] = mole / total volume = 8/4 = 2 M [K+] = 6/4 = 1.5 M [Cu2+] = ¼ = 0.25 M
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