now b/c M = moles / L to get the answer in g / L subs what you get for x into this: x moles / L * [235.77 g / mol] = answer in g/L. Hint -- First Convert 0.0072 G/L To Mol/L So You Can Use ICE. 0 0. What mass of AgBrO3 can be dissolved in 50.0 mL of 0.500M NaBrO3 at 80°C? The solid phases of aqion are listed here in two tables (together with the solubility product in form of pK sp = - log 10 K sp):. Calculate Ksp. The Ksp for AgBrO3 is 4.00 x 10^-2. x = √Ksp M. Molar mass AgBrO3 = 107.87 + 79.90 + 16.00 * 3 = 235.77 g/mol. Discuss the results obtained in (i) and (ii). ii. Calculate the solubility of AgBrO3 in a 0.1 mol/kg AgCl (aq) solution at 25oC. Get your answers by asking now. You’ll find ksp data in the picture so Ksp = x². (MW Of AgBrO3 Is 236 G/mol). I got 9.43 x 10^-2 g. Chemistry Question: The Solubility Of Silver Bromate, AgBrO3, In Water Is 0.0072 G/L. Table sorted by formula; Table sorted by mineral name; Only a subset of these minerals enter the equilibrium calculations by default. Answer to: Calculate the solubility of AgBrO3 in water at 25c. Solubility Product Constants K sp at 25°C. (0.16 g AgBrO3 / (235.7704 g AgBrO3/mol)) / (0.100 L) = 0.0068 mol/L Calculate the solubility of AgBrO3 in water at 25oC considering the activity coefficient of the ions obtained using Davies Equation. Still have questions? The source below says the solubility of AgBrO3 at 20°C is 0.16 g/ 100 mL H2O. i. iii. Ask Question [1ΔS f (Ag+1 (aq)) + 1ΔS f (BrO3-1 (aq))] - [1ΔS f (AgBrO3 (s))] [1(72.68) + 1(163.18)] - [1(151.9)] = 83.96 J/K 83.96 J/K (increase in entropy) Ksp for AgBrO3 in water at 25 oC and 1 bar is 5.38 x 10-5 mol2 /kg2 .
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