In such a case together with, we are able to expose a love between your K

There is no such tendency shown by Ct and therefore [H + escort in Peoria ] > [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 K_b as K_h.K_b = K_w Let us calculate the Kb value in terms of degree of hydrolysis (h) and the concentration of salt K_h = h dos C and

Question 19. The solubility product of Ag₂CrO_cuatro is 1 x 10 -12 . What is the solubility of Ag₂CrO₄ in 0.01 M AgNO₃ solution? Answer: Solubility product of Ag₂CrO₄, K_sp = 1 x 10 -2

Matter twenty six

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, K_sp = [Ca 2+ ] 3 . [PO₄ 3- ] 2 K_sp = (3s) 3 . (2s) 2 K_sp= (27 s 3 ) . (4s 2 ) K_sp = 108s 5 .

Precipitation of 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 K_spof CaF₂? Answer: CaF_{dos (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. K_sp 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 K_sp = [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) K_sp = [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 > K_sp [Pb 2+ ][Cl – ] 2 = (0.06)(0.08) 2 = 3.84 x 10 -4 Since ionic product [Pb 2+ ][Cl – ] 2 > K_sp 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) K_sp = [Al 3+ ] [OH – ] 3 Al(OH)₃ precipitates when [Al 3+ ] [OH – ] 3 > K_sp (1 x 10 -3 )[OH – ] 3 > K_sp [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