Calculate the equilibrium concentration of Zn 2+ ion in a solution prepared by d

Question

Calculate the equilibrium concentration of Zn2+ ion in a solution prepared by dissolving 1.0 g of ZnCl2 and 8.1 g of NaOH in enough water to produce 1.0 L of solution. How / show work.

8.7E-17

2.9E-15

1.6E-15

4.0E-13

2.4E-21 Calculate the equilibrium concentration of Zn^2+ ion in a solution prepared by dissolving 1.0 g of ZnCl2 and 8.1 g of NaOH in enough water to produce 1.0 L of solution. How / show work. Zn^2+ + 4 OH^- [Zn(OH)4]^-2 For [Zn(OH)4]^2-, Kf = 2.8 x 10^15. 8.7E-17 2.9E-15 1.6E-15 4.0E-13 2.4E-21

Explanation / Answer

mole of ZnCl2 = wt/M.wt = 1/136.4 = 0.00733 mole of NaOH = Wt/M.wt = 8.1/40 = 0.2025 Zn2+ + 4 OH- -----------> [Zn(OH)4]2- initial 0.00733 0.2025 0 Equilibrium 0 (0.2025-4*0.00733) 0.00733 = 0.173 Now after formation of complex lets say x amount of complex dissociates to yield ions, hence equilbrium after change x is x 0.173 + 4x 0.00733-x now formation constant Kf = (0.00733-x) / x * (0.173+4x)^4 since x is very small we can rewrite it as Kf = 0.00733/ x * 0.173^4 on solving u get x = 2.9 * 10^-15 hence 2nd choice is the answer

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