Question

Question 1 For the aqueous (Hg(NH3)4]2+ complex has a Kt - 1.8 x 1019 at 25°C. Suppose equal volumes of 0.0082 M Hg(NO3)2 solution and 0.72 M NH3 solution were mixed. Calculate the equilibrium concentration (Mof aqueous Hg2+ ion. (Please enter your answer is scientific notation. For example if you answer was 3.15 x 106, please enter 3.155) Answer should be in 2 significant figures

Answer 1

Genen: kfz 168X1019 tempr=25°c concentration = 0.0082m of highwaz, solution concentration solution of NH3 0.72M solution: Suppose that equal volume v Total volume of final solution a VAV 22V 28 Hghe + uning - Engenuz?u?? Imole 4moles Imole moles of night (Mx V) 20.0082 x V moler Moles of N3 20.72v mole. 1 mole 4 mole niz geres complex 0.72V moles uz → gives moles complex = 0.72V Mole of complex Nuz t CS Scanned with CamScanner 20.18v can give

complex - moles complexe Total volume 20.0082v moles 2v 2o.Doul moles [NH] in excess Tosal moles-Moles reached Tosal volume z 0.720-4 (0.00821) 2v 2 0.72-0.0328 av 70.3436 M. Kp [ng cwus)??? [Mg2+] (NH 3] 2 0.0041 [ngat) (1-8x10!)x(0.3436)4 [right] = 0.004) 1.8x1019 x 0.0139 0.004) 2) 0.164X 1079 0.025X1019 CS Scann[ng2 trachiherdoner -20 PM