Question

5. Please calculate the pH of a solution containing 0.40 M CH,COOH and 0.70 M CH3COONa (K, acetic acid = 1.8 X 10"): CH3COOH + H+ + CH3COO CH3COONa → CH3COO + Nat 6. The molar solubility of tin(ll) iodide is 1.28 x 10-2 mol/L. What is Ksp for this compound?

Answer 1

6.5: We know from Henderson equation, the formula of bn of buffer solution - [salt pH = bka + log - 9 [Acid] Where, ka = ionisation constant of Acid [Soft] = Concentration of salt [Acid] = concentration of Acid Here solution containing CH₂COOH and CH ₂ COONa. That is the solution in a buffer. Ka of CH3COOH = 3:8x10-5 [ Sost] = [c Hz Coona] = 0.70 M [Acid] = [CH3COOH] = 0.40 M

Accordingly, ÞH = $ka + dog [CH3COONa ] 4 Сенасоо н = -dog (ka) + dog 60 70 = - 10g (118x10-5) + dog on the = 4.74 + 0.24 = 4.98 in pr of the solution in 4.98. Am.

He have, Tin (11) iodide → Sn Iz Sn 12(0)5 su2+ (0g) + 2 I- (aq) given, Molan solubility of Sn. Iz is 1.28x 10-2 mol/L. So from Sn Ig , One S n2+ ion and two I ion will be produced. So, ' : [S12+] = 128 x 10-2 mol/L [I-] = (4.28 x 10-2 mol/L) x2 = 2.56 x 10-2 mol/L : Ksp (Solubility product) = [ Sn2+] x [9-12 = [128x10-2] * [2.56 X10-232 = 8.39 x10-6 . Ksp of Sn Iz is 8.39 x 10-6. Ana.