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Assuming the complete dissociation of HCl and the lead salt ,calculate how much HCl is to be added to a 0.001 M lead salt solution to just prevent precipitation when saturated with H2S?The concentration of H2S in its saturated solution is 0.1 M. KaH2S=1.1×10-23;KspPbS=3.4×10-28.
Hi Ankur, [Pb+2] [S2-] = Ksp = 3.4 * 10^-28 [PbS] = [Pb+2] = 0.001 (saturated) [S2-]= 3.4 * 10^-25 Ka = 1.1 * 10^-23 = [H+]2[S-2]/[H2S] [H+]2 = 1.1/3.4 * 10^2 * 0.1 [H+] = 1.8 since due to the very small Ka value of H2S the H+ furnished by it is negligible. Therefore the 1.8 M HCl is to be added.
Hi Ankur,
[Pb+2] [S2-] = Ksp = 3.4 * 10^-28
[PbS] = [Pb+2] = 0.001 (saturated)
[S2-]= 3.4 * 10^-25
Ka = 1.1 * 10^-23 = [H+]2[S-2]/[H2S]
[H+]2 = 1.1/3.4 * 10^2 * 0.1
[H+] = 1.8
since due to the very small Ka value of H2S the H+ furnished by it is negligible.
Therefore the 1.8 M HCl is to be added.
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