Question

18. A 200.0-mL sample of a 0.0015 M copper(II) nitrate solution is mixed with a 250.0-mL sample of 0.20 M NH3. After the solution reaches equilibrium, what concentration of copper(II) ion remains? You will have to look up the appropriate complex formation constant.

Answer 1

I Given zoome of. Cu(NO3)2 0.08ISM (CaCO3)2) = 200 mL x 0.0015 19 = 0.3 mmdes 250mL q NH₃ q 0.209 - [NH3] = 250mLX 0.209 = (NH3] = 50 mmdes 27 2 of Ca CNH3)A HUU Cu2+ + 4NH3 6. Emo + mm -x -40 0.3-2 50-4% ta formatim constant from the leterature, Kf = complex Kf = 1x103

{ca CN Hz)?] KG [cu 27] [NH3)4 » 1111x10'3 = x (0.3-2) (350-4274 as denominator is high, so we can neglect (68.4077 1.1 *10*2= (3-2) → = a wwx1013 (0.3-7)= x 0.33x10'3 - 1.1x1013xzx. 0.33 x103 = (1+111x10'3 ) - 0.2973'm moles 3a 0.33x8013 loll trots - ..^. the norgmoles of cu 2+ 'at Equilibrium = [Cu2+] = 0.32 = 0.3-02973 = 0.0027 m moles = 2.7x10 moles.

..". The concentration q [Ca 2+] at Sauilibrium z noi qmoles . Total volume - 2-7x10 6 moles - looo asom Ecuat]= 6.410619

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