Question

5. A solution is made by mixing 200.0 ml of 1.5 x 10-4 M Cu(NO3)(aq) with 250.0 ml of 0.20 M NH3(aq). Calculate the concentration of copper ion (Cu? (aq)) in the solution when it reaches equilibrium. (Kr for [Cu(NH3)4]2+ = 1.7 x 1013)

Answer 1

Given, Concello De Initial concentration of Cu (NOZ) = 1.5x10 ²M = 1.5xlo mol/L Volume of Cu(Nola soluchons used 3 200.0 mL Initial males of Culnoza = Concentration X Volume 105x103 mol/Lx200.0m2 = 0.30 mmol Initial concentration of Allg = 0.20 M = 0.20 mol/L Volume of Nity Solution cesed 2 250.0 ml Initial moles of NH3 concentration & Volume 0.20 mol/Lx250,0 mL 50.0 mmol 2+ Jonic equation for the reachion is Custrag) + 4 Ntg Cep) -> (cu CN1 H24] cap) Initial 0.30 monol 50.0 mmol After readhon 0 mmol 50.0-4x0.30 = 48.8 mmol 0.30 mmol x mmel At equilibrium moles At equilibrium Concentration (48.8 + x) mmol (48.8+x) Am 450.0 (0.30-x)mmal (0.30-x) 450.0 Note - Total Volume of solution - 450.00 do 450.0 2000F 2500 a

24 Now [Cu(NH3)² [ano] [NH2J4 Given, Kr for [cu CN Hz q7 200 167x 2013 Putting all the values in above equation, we get 1.7x1013 = Ce sord 0.30-X 450.0 48.872 14 (450.0) 450.0 Suppose 2 < 0.30 or 48.8 So + 1o7x10'3 0.30 (450.0 lo so..) (4800 48.8 150.0 t456.0 Earth 450.0 (6.67X104) (1.7 x 108) (0.10834 2.835x10 M ~ 2.8 x 10-13 M -13 (astma 450.0 Hence, the concentration of copper ion Captcaq)) in the solution when equilibrium reaches 98 2.821018M