a)
At equilibrium:
AgCl <----> Ag+ + Cl-
s s
Ksp = [Ag+][Cl-]
1.6*10^-10=(s)*(s)
1.6*10^-10= 1(s)^2
s = 1.265*10^-5 M
Molar mass of AgCl,
MM = 1*MM(Ag) + 1*MM(Cl)
= 1*107.9 + 1*35.45
= 143.35 g/mol
Molar mass of AgCl= 143.35 g/mol
s = 1.265*10^-5 mol/L
To covert it to g/L, multiply it by molar mass
s = 1.265*10^-5 mol/L * 143.35000000000002 g/mol
s = 1.813*10^-3 g/L
Answer: 1.8*10^-3 g/L
b)
At equilibrium:
AgBr <----> Ag+ + Br-
s s
Ksp = [Ag+][Br-]
5*10^-13=(s)*(s)
5*10^-13= 1(s)^2
s = 7.071*10^-7 M
Molar mass of AgBr,
MM = 1*MM(Ag) + 1*MM(Br)
= 1*107.9 + 1*79.9
= 187.8 g/mol
Molar mass of AgBr= 187.8 g/mol
s = 7.071*10^-7 mol/L
To covert it to g/L, multiply it by molar mass
s = 7.071*10^-7 mol/L * 187.8 g/mol
s = 1.328*10^-4 g/L
Answer: 1.3*10^-4 g/L
c)
At equilibrium:
Ag2CrO4 <----> 2 Ag+ + CrO42-
2s s
Ksp = [Ag+]^2[CrO42-]
9*10^-12=(2s)^2*(s)
9*10^-12= 4(s)^3
s = 1.31*10^-4 M
Molar mass of Ag2CrO4,
MM = 2*MM(Ag) + 1*MM(Cr) + 4*MM(O)
= 2*107.9 + 1*52.0 + 4*16.0
= 331.8 g/mol
Molar mass of Ag2CrO4= 331.8 g/mol
s = 1.31*10^-4 mol/L
To covert it to g/L, multiply it by molar mass
s = 1.31*10^-4 mol/L * 331.8 g/mol
s = 4.348*10^-2 g/L
Answer: 4.3*10^-2 g/L
please show how you get an answer please 1. Determine the solubility in g/L of each...
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ANSWER ALL PARTS 1. Determine the molar solubility of AgBr in a 0.50 M NH3 solution. The Ksp for AgBr is 5.0 x 10-13 and the Kf for Ag(NH3)2+is 1.7 x 107. 2. How many unpaired electrons are there in the Zn2+ ion in the Zn(NH3)62+ complex ion? 3. Which of the following complex ions will absorb the shortest wavelength of visible light? Ni(CN)64- Ni(NH3)62+ Ni(H2O)62+ NiCl64-
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Determine the molar solubility of AgBr in a solution containing 0.250 M MgBr2 (strong electrolyte). Ksp (AgBr) = 7.7 x 10-13 O A.8.8 x 107 M B.5.1 x 10-12 M c. 1.5 x 10-12 M O 0.5.8 x 10-5 M E.0.150 M
Please show work
Consider the following Gibbs energies at 25 degree C. calculate delta G degree _rxn for the dissolution of AgCl(s). Calculate the solubility-product constant of AgCl. Calculate delta G degree_rxn for the dissolution of AgBr(s). Calculate the solubility-product constant of AgBr.
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solubility of the slightly soluble salt The solubility of AgCl is the same in a 1.0 M NH, solution vs. in pure water because the Ksp of AgCl is a constant at 25°C. D) 9. Calculate the molar solubity of AgBrts) in 0.500 M NH, at 25°C. Kup of AgBr is 1.8 x 10-5, K of Ag(NHs)2" is 1.7 x 10 A) 8.7 x 104 M B) 1.2x 10-3 M C) 4.2 x 10-7 M...