In such a case in addition to, we are able to expose a relationship between the K

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In such a case in addition to, we are able to expose a relationship between the K

There is no such tendency shown by Ct and therefore [H + ] > [OH – ] the solution is acidic and the pH is less than 7. As discussed in the salt hydrolysis of strong base and weak acid. a and Kb as Kh.Kb = Kw Let us calculate the Kb value in terms of degree of hydrolysis (h) and the concentration of salt Kh = h dos C and

Question 19. The solubility product of Ag2CrOcuatro is 1 x 10 -12 . What is the solubility of Ag2CrO4 in 0.01 M AgNO3 solution? Answer: Solubility product of Ag2CrO4, Ksp = 1 x 10 -2

Concern 26

Question 20. Write the expression for the solubility product of Ca3(PO4)2 Answer: Ca3(PO4)2 (s) \(\rightleftharpoons\) 3Ca 2+ (3s) + 2PO4 3- (2s) The solubility of Ca3(PO4)2 is, Ksp = [Ca 2+ ] 3 . [PO4 3- ] 2 Ksp = (3s) 3 . (2s) 2 Ksp= (27 s 3 ) . (4s 2 ) Ksp = 108s 5 .

Precipitation regarding PbCI

Question 21. A saturated solution, prepared by dissolving CaF2(s) in water, has [Ca 2+ ] = 3.3 x 10 -4 M. What is the Kspof CaF2? Answer: CaFdos (s) \(\rightleftharpoons\) Ca 2+ (aq) + 2F(aq) [F] = 2 [Ca 2+ ] = 2 x 33 x 10 -4 M = 6.6 x 10 -4 M = [Ca 2+ ] [F – ] 2 = (3.3 x 10 -4 ) (6.6 x 10 -4 ) 2 = 1.44 x 10 -10

Question 22 Santa Ana escort. Ksp of AgCl is 1.8 x 10 -10 . Calculate molar solubility in 1 M AgNO3 Answer: AgCl(s) > Ag + (aq) + Cl – (aq) x = solubility of AgCl in 1M AgNO3

[Cl – ] = x Ksp = [Ag + ] [Cl – ] 1.8 x 10 -10 = (1) (x) x = 1.8 x 10 -10 M

Question 23. A particular saturated solution of silver chromate Ag2CrO4 has [Ag + ] = 5 x 10 -5 and [CrO4] 2- = 4.4 x 10 M. What is the value of for Ag2CrO4? Answer: Ag2CrO4 (s) \(\rightleftharpoons\) 2Ag + (aq) + CrO4 2- (aq) Ksp = [Ag + ] 2 [CrO4 2- ] = (5 x 10 -5 ) 2 (4.4 x 10 -4 ) = 1.1 x 10 -12

Will a precipitate be formed when 0.150 L of 0.1 M Pb(NO3)2 and 0.100 L of 0.2 M NaCl are mixed? (PbCI2) = 1.2 x 10 -5 . Answer: When two or more solutions are mixed, the resulting concentrations are different from the original. Total volume 0.250L

2 (s) occurs if [Pb 2+ ][Cl – ] 2 > Ksp [Pb 2+ ][Cl – ] 2 = (0.06)(0.08) 2 = 3.84 x 10 -4 Since ionic product [Pb 2+ ][Cl – ] 2 > Ksp PbCl2 is precipitated.

Question 27. of Al(OH)3 is 1 x 10 -15 M. At what pH does 1.0 x 10 -13 M AI 3+ precipitate on the addition of buffer of NH4CI and NH4OH solution? Answer: Al(OH)3 ? Al 3+ (aq) + 3OH – (aq) Ksp = [Al 3+ ] [OH – ] 3 Al(OH)3 precipitates when [Al 3+ ] [OH – ] 3 > Ksp (1 x 10 -3 )[OH – ] 3 > Ksp [OH – ] 3 > 1 x 10 -12 [OH – ] > 1 x 10 -4 M [OH – ] = l x 10 -4 M pOH = – 1og10[OH – ] = – log (1 x 10 -4 ) = 4 pH = 14 – 4 = 10 Thus, Al (OH)3 precipitates at a pH of 10

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