Ksp Zn(OH)2 = 7.7 x 10 ^ - 17 50 ml of .1M Zn(NO3)2 is mixed with 50.0 ml of .3M

Question

Ksp Zn(OH)2 = 7.7 x 10 ^ - 17 50 ml of .1M Zn(NO3)2 is mixed with 50.0 ml of .3M NaOH Calculate molar concentration of ZN+2 in resulting solution once equilibrium has been established. Assume volumes are additive

Explanation / Answer

The value of the solubility-product constant, Ksp, for Zn(OH) 2 is 7.7 x 10¯ 17 at 25°C. (i) Calculate the solubility (in moles per liter) of Zn(OH) 2 at 25°C in a solution with a pH of 9.35. (ii) At 25°C, 50.0 milliliters of 0.100-molar Zn(NO3) 2 is mixed with 50.0 milliliters of 0.300-molar NaOH. Calculate the molar concentration of Zn2+(aq) in the resulting solution once equilibrium has been established. Assume that volumes are additive answer:(i) pH =9.35 ----> pOH = 4.65 ----> [OH¯ ] = 2.24 x 10¯ 5 M (one point) [Zn2+] = Ksp / [OH¯ ]2 = 7.7 x 10¯ 17 / (2.24 x 10¯ 5)2 = 1.5 x 10¯ 7 M (one point) One point earned for correct determination of [OH¯ ] One point for correct answer (assume [Zn2+] equals solubility in moles per liter) No points earned if [OH¯ ] is assumed equal to twice [Zn2+] (ii)Zn(OH)2(s) ---> Zn2+ + 2 OH¯ (0.050-x (0.150-2x) Ksp = 7.7 x 10¯ 17 = [Zn2+] [OH¯ ]2 = (0.050 - x)(0.150 - 2x)2 (one point) Solve for x, then subtract x from 0.050 M to obtain [Zn2+] (one point)

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