Question

1.Given a solution in which [Cu²⁺] = 0.20 mol/L and [NH³] = 1.40 mol/L, what is the equilibrium concentration of free Cu²⁺? The complex Cu(NH3)4²⁺ has Kf = 5.0 X 10¹³. The answer to the question is 3.1 X 10 ^-14 I just need how to do it with the work shown.

2. You titrate a 20.00 mL sample of ammonia (Kb = 1.8 X 10^-5) to the equivalence point using 10.00 mL of .20 M HCl solution. What is the pH of the solution at the equivalence point of the titration. The answer to the question is 5.21 but I do not know how to get there.

Answer 1

Bolution: Ô olution [Cu?'] = 0.2 molle [Nm] = 1.40 mol/L 2 . Cu toa) + 4 NMD lens = [Cu (N12)4] : Ef = 5-oxiol Limiting Reagent = Cy?t because its mode is less than. (Nh3 ] Since the Formation constant value is very ligh it means reaction is almost completed. So that we can take the value of n. as 0.2 Oil : 1.40 0 Cu2+ + 4 N = [cu (Nm3) 47²+ 012-11 104-48 & BL taking n=0.2 because the value of kf is high 0 - 1.4-0.8 - 0-6 ore : K = [eu lwng)]? (642+] [ nns] 0.2 zo2 [ca2+] (0.674 - (2477) 01296 CS Scanned with . [Cu²+] = soit aux 0-1296 +12 50 X 10 = . CamScanner

Icy (cy ²+] =...0.2 J = 0.648 X10'S Trdi TTHC [C42+] = 1 Cu2+] = (C4) = 0.3086 X1013 3.08 xron!4 3.1 Xroly Answer M solution ③ Total volume of solution at oquivalence point = 20+ 10 = 30ml. poma at équivalence point marles of NH3 = moly of nl = males of salt = lomexo.am = 2 mmol mody of salt = 2 mmal Concentration of salt (Nhuch) = males - Total volume . = 2m mol Concentrati . Bomb. . .. = 0.0667(hos The ph of П 2. salt is given by ри = 1 - Jej c -; • PRь 2 whese c= concentration of . pkb = -doy Kb = . - log. (1.8X105) pkn = 4.74 salto cs Scanned with CamScanner

ph=7-12 log (60667) - 12 X 4.74 P) = 1 - x (-1-12t) - 237 (pn = 5.21) Angweni Scanned with CamScanner