Question

At 25 °C, only 0.0450 mol of the generic salt AB is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? AB(s) 7 A+ (aq) +B+ (aq) Ksp =

Answer 1

n = 0.0450 mol, V IL C = n = 0.0450 mol = 0.048om/L A ABE) g) + Bien 00410 o ouro couro CSP Ksp = d.ouro xo. 0450 = 2.025x10-31 M(OH), mat + 2015 pH of the soleation = 10.10 poth of the solution = 3.82 [H] = 10 -3.82 = 1513x104 [on] = 1.513 x 10-le t on Ksp = [m²+ ] [online = 1,513 x10.4 x 1.513 x 10-4 Win 0 1567x10 4 x 1.813 8104 1 ksp = 1.145810-8

Answer 2

# moles of salt AB = 0.045 mol voluone of ceeater = 1L mores of solute (43) concentration (or molarity) = Ang volume of solutions line) 0.045 mes = 0.095 mol IL 2 = 0.045M [AB] = 0.045m & Co.

Date now, AB(C) = A+ coug) + B-lag). Initially to Со After Bissociation Ksp = 2 [A+] [ 6-3 (6)x(60) = (0.04509)Y (0.045 = 2.0258103 Kepa 2.025 X 10 3 pot = longo 10-18 other to photpot 214 14-10.18 3.82 por= 3.82 port = -log [on- [ost-} = 10-3.82 m.

[он-1 : 10- 3-62 м е м ін (9) + 2 он(49) м(он), Initiallep: c finally oo [ои-2 - (ас) = 10 3 2 2c: a lo-3-83 Саоѕ х 10 3gp - Гw2+2 х Гон-1- (c) х (2c) 2- 4c3 . s (4 (ors ri-32 Р 2 - о:s x 1 146 A S X 10- 12.46 ksp = S X16-12- 44