Calculate the solubility at 25 °C of AgCl in pure water and in a 0.0170 MAgNO3 s

Question

Calculate the solubility at 25 °C of AgCl in pure water and in a 0.0170 MAgNO3 solution. You'll find Ks, data in the ALE Formula BaCrO4 BaSO4 CaCO3 CaF2 Co(OH)2 CuBr CuCO3 Fe(OH)2 sp 1.17x10-10 Data tab 1.08x10-10 3.36×10-9 3.45x 10-11 5.92x10-15 6.27x109 1.4x10-10 4.87x10-17 7.40x10-14 2.8x10-13 3.3×10-8 5.61×10-12 5.48x10-16 5.38×10-5 8.46 × 10-12 1.77x10-10 1.12x10-12 5.60x10-10 1.46x 10-10 3.0x10-17 5.35x10-13 1.77x10-10 Round both of your answers to 2 significant digits solubility in pure water: solubility in 0.0170 M AgNO3 solution: PbCrO4 PbF2 Mg(OH)2 Ni(OH)2 AgBrO3 A92C03 Agcl Ag2Cr04 SrCO3 ZnCO3 Zn(OH)2 AgBr AuCl

Explanation / Answer

AgCl ===== Ag+ + Cl-

a) Ksp in pure water = [Ag+][Cl-]

1.77 * 10^(-10) = s * s

s^2 = 1.77 * 10^(-10)

s = 1.3304 * 10^(-5)

Solubility in g/L = 1.3304 * 10^(-5) * Molar mass of AgCl

=> 1.3304 * 10^(-5) * 143.32

=> 1.9067 * 10^(-3) g/L

AgCl ===== Ag+ + Cl-

b) Ksp og AgCl = [Ag+][Cl-]

1.77 * 10^(-10) = (s+0.0170)(s)

assuming s is very small in comparison to 0.0170 we can write

s= 1.77 * 10^(-10)/0.0170 = 1.0411 * 10^(-8)

Solubility in g/L = 1.3304 * 10^(-5) * Molar mass of AgCl

=> 1.0411 * 10^(-8) * 143.32

=> 1.4921 * 10^(-6) g/L

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