(Ksp= 5.0*10^-13) AgBr
A) calculate the molar solubility of AgBr in pure water.
B) calculate the molar solubility of AgBr in 2.8×10^−2 M AgNO3 solution.
C)calculate the molar solubility of AgBr in 0.11 M NaBr solution.
2 Answers
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5.0 x 10^-13 =(x)(x)
x = molar solubility = 7.1 x 10^-7 M
5.0 x 10^-13 = (x + 2.8 x 10^-2)(x)
x = molar solubility = 1.8 x 10^-11 M
5.0 x 10^-13 = (x)(0.11+x)
x = molar solubility =4.5 x 10^-12 M
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Hydroxyapaptite is, more accurately call hydroxylapitate. Ca5(PO4)3(OH) So the solubility product is = Ksp = [Ca2+]^5 x [PO4 3-]^3 x [OH-] let the molar solubility of hydroxyapapite = a with units mol/dm3 so Ksp = (5a)^5 x (3a)^3 x a = 84375a^5 x 27a^3 x a Ksp = 3125a^5 x 27a^3 x a Ksp =84375x a^9, rearranging a = 9√(Ksp / 84375)) so a = 2.8355e-5 molar solubility to 1.s.f. = 3e-5 moles per dm3 or 30 micromolar.
