Question

In the presence of excess OH^-, the Zn²⁺(aq) ion forms a hydroxide complex ion, Zn(OH)₄^2-. Calculate the concentration of free Zn²⁺ ion when 1.54×10^-2 mol Zn(CH₃COO)₂(s) is added to 1.00 L of solution in which [OH^- ] is held constant (buffered at pH 12.70). For Zn(OH)₄^2-, K_f = 4.6×10¹⁷.

[Zn²⁺] =  M

Answer 1

For a buffer solution of oft = 12.70, the concentration of ot in can be calculated as Pollows - HH 2 12-70 POH 2 14 -pott . 14-12-70 or, DOH = 1:30 = -log(04) - 1.20 29 POH-] á 10-130 151) POH-] = 0.05 m Now Concentration 2024 in, ist adked , : 1221) ; moles of th (CH3 Coo) volume of solution & 1.54x162 mot 1.00L . = 1.59x102 mol/L = 1:54x102 M

As LOH] » [2224] hence the addition of 2n (C208)2 in the buffer solution, will formi hydroxide complex in , ?n 20H)472- as follows. 2021 7 4 OH Zn(OH) 72 - The formation constant, In con Rf [229] corra Since all 2221 forms 2n(874) 2. ith complex in presence of excess of ions. Hence [21601942.) – [2n"/add [Zn(OH)] = 1.54.8102M .:. [2421) - [ Z1(04)*7 : 9 . Hence xf * [ot] 1.59x102 2 [2429) = 4.681017x Co-054 ... [Zn2+] . 5.36.x105 m