Question

A generic salt, AB,, has a molar mass of 307 g/mol and a solubility of 6.40 g/L at 25 °C. What is the Ksp of this salt at 25 °C? AB3(s) = A3+ (aq) + 3B- (aq) Ksp = 5.25 x10-6

Answer 1

The molar mass of - = 6.407 flitre molar mass of the salt ABs is = 301J/mole Solubility of salt Then, kse = ? Cat 25°c) and molar Salubility ksp : 52.5X10° (gives) and Kse of Abg = ? Thus, we have - - ABZ.CS) = A Caqu) +38:'(aqua) 2 The expression of kse = [A3+] [B-39 Now, the stoichiometry of equatio Shows a the relationship Thes, for [A3f] and [Bo] Kse will be [A3+] = 3x(5:25X10 ) = 15.25x10-6 M [63) = 5.25x1064 © Putting, these values in Ksp espression = - Ksp = [15. 75X10% ][ 5.25x10 = [15375 XL2 I (144.70 X10%) [:57x105] [1:44 EXO - 20278 1021) The Ksp of generic salt_abz will be [20237104"}