write-a-balanced-net-ionic-equation-for-each-of-the-following-dissolution-reacti

Question

What is the net ionic equation for the dissolution of Cu(OH)_{2} in NH_{3}? K_{sp} for Cu(OH)_{2} is 1.60×$$10^{–19}$$ and K_{f} for [Cu(NH_{3})_{4}]^{2+} is 5.60×$$10^{11}. $$Determine the equilibrium constant for the dissolution reaction.

Accepted Answer

Cu(OH)_{2}(s) + 4NH_{3} (aq) ⇌ $$Cl^{–}(aq$$) + [Cu(NH_{3})_{4}]^{2+}(aq) K = 8.96×$$10^{–8}$$

Answer

Cu(OH)_{2}(aq) ⇌ $$Cu^{2+}(aq) + 2$$ $$OH^{–}(aq)(aq) K = 4.96$$×$$10^{–8}$$

Answer

Cu(OH)_{2}(aq) + 4NH_{3} (aq) ⇌ $$Cl^{–}(aq$$) + [Cu(NH_{3})_{4}]^{2+}(aq) K = 4.41×$$10^{–9}$$

Answer

Cu(OH)_{2}(s) ⇌ $$Cu^{2+}(aq) + 2$$ $$OH^{–}(aq)(aq) K = 2.18$$×$$10^{4}$$

What is the net ionic equation for the dissolution of Cu(OH)2 in NH3? Ksp for Cu(OH)2 is 1.60×10–19 and Kf for [Cu(NH3)4]2+ is 5.60×1011. Determine the equilibrium constant for the dissolution reaction.

What is the net ionic equation for the dissolution of Cu(OH)₂ in NH₃? K_sp for Cu(OH)₂ is 1.60×10^–19 and K_f for [Cu(NH₃)₄]²⁺ is 5.60×10¹¹. Determine the equilibrium constant for the dissolution reaction.