Chem

1a) Calculate submarine sandwich solubility of SrF2 in: 0.010 M Sr(NO3)2 SrF2 <=> Sr+2 & angstrom; 2 F- Ksp = [Sr+2] [F-]^2 4.3x10^-9 = [0.010 M] [F-]^2 [F-]^2 = (4.3x10^-9) / [0.010 M]  [F-]^2 = 4.3x10^-7 [F-] = 6.56 X 10^-4 sub by the compare: 1 groyne of SrF2 <=> Sr+2 & worldwide ampere; 2 moles of F- the bomber solubility of SrF2 that releases [F-] is half as much: 3.279 X 10^-4 hero SrF2 yoiur answer, travel to 2 sig figs would be 3.3 X 10^-4 Molar SrF2 ====================================== 1b) Calculate molar(a) solubility of SrF2(Ksp=4.3x10^-9) in 0.010 M NaF Ksp = [Sr+2] [F-]^2 4.3x10^-9 = [Sr+2] [0.010]^2 [Sr+2] = 4.3x10^-9 / [0.010]^2 [Sr+2] = 4.3x10^-9 / 1 X 10^-4 [Sr+2] = 4.3x10^-5 Molar by the equation: 1 mole of SrF2 <=> 1 mole of Sr+2 & 2 F- the Sr+2 in termination comes from an equal molar solubility of SrF2 molar solubility of SrF2 = 4.3x10^-5 Molar ========================== ============ 2) Calculate the molar solubility of Agl in 0.10 M NaCN. Kf for (Ag(CN)2-) is 3.0x10^20 ; Ksp for Agl is 8.5x10^-17 Agl <=> Ag+ & I- Ksp = [ Ag+] [I-] Ag+ & 2 CN- <=> [Ag(CN)2]- Kf = [Ag(CN)2]- / [Ag+] [CN- ]^2 Knet = (Ksp) (Kf) = [ Ag+] [I-] times [Ag(CN)2]- / [Ag+] [CN- ]^2 Knet = (8.5x10^-17) (3.

0x10^20) = [I-] [Ag(CN)2]- / [CN- ]^2 2.55 X 10^4 = [I-] [Ag(CN)2]- / [CN- ]^2 when the CN- added increases the solubility of Agl  the equilibrium: AgI <=> Ag+ & I- theoretically let go an equal mensuration of Ag+ & I- but when it shifts to the right, because the 0.1 Molar CN- takes out the Ag+ forming [Ag(CN)2]- we assume that essentially all of the Ag+ released is! converted into [Ag(CN)2]- so, preferably of producing an equal kernel of Ag+ & I- it produces an equal amount of [Ag(CN)2]- & I- which we refer to as X returning to: 2.55 X 10^4 = [I-] [Ag(CN)2]- / [CN- ]^2 becomes: 2.55 X 10^4 = [X] [X] / [0.10 M ]^2 X^2 = (2.55 X 10^4) [0.10 M ]^2...If you want to get a full essay, raiment it on our website: OrderEssay.net

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