Barium iodate (Ba(IO3 ) 2) is sparingly soluble in water (Ks = 1.5×10-9 ).i) Calculate the solubility (g L-1 ) of Ba(IO3 ) 2 in pure water. [3 marks]ii) Calculate the molar concentration of Ba2+ and IO3- at the equilibrium in purewater. [2 marks]iii) Calculate the pH of a solution of Ba(IO3 ) 2 in pure water at the equilibrium(Ka(HIO3 ) = 1.7×10-1 ). [2 marks]iv) Calculate the solubility (g L-1 ) of Ba(IO3 )2 in an aqueous solution of BaCl 20.01 M. [4 marks]v) Calculate the molar concentration of Ba2+ and IO3- at the equilibrium in anaqueous solution of BaCl 2 0.01 M. [2 marks]vi) Calculate the pH of a solution of Ba(IO3 ) 2 in an aqueous solution of BaCl 20.01 M at the equilibrium (Ka(HIO3 ) = 1.7×10-1 )
Question
Barium iodate (Ba(IO3 ) 2) is sparingly soluble in water (Ks = 1.5×10-9 ).i) Calculate the solubility (g L-1 ) of Ba(IO3 ) 2 in pure water. [3 marks]ii) Calculate the molar concentration of Ba2+ and IO3- at the equilibrium in purewater. [2 marks]iii) Calculate the pH of a solution of Ba(IO3 ) 2 in pure water at the equilibrium(Ka(HIO3 ) = 1.7×10-1 ). [2 marks]iv) Calculate the solubility (g L-1 ) of Ba(IO3 )2 in an aqueous solution of BaCl 20.01 M. [4 marks]v) Calculate the molar concentration of Ba2+ and IO3- at the equilibrium in anaqueous solution of BaCl 2 0.01 M. [2 marks]vi) Calculate the pH of a solution of Ba(IO3 ) 2 in an aqueous solution of BaCl 20.01 M at the equilibrium (Ka(HIO3 ) = 1.7×10-1 )
Solution
i) The solubility product expression for Ba(IO3)2 is Ksp = [Ba2+][IO3-]^2. Since it dissociates into one Ba2+ ion and two IO3- ions, we can say that [Ba2+] = s and [IO3-] = 2s. Substituting these into the Ksp expression gives Ksp = s*(2s)^2 = 4s^3. Solving for s (the molar solubility) gives s = (Ksp/4)^(1/3) = (1.5×10^-9/4)^(1/3) = 6.22×10^-4 M. To convert this to g/L, multiply by the molar mass of Ba(IO3)2 (487.2 g/mol) and by 1000 L/m^3 to get 303 g/L.
ii) As stated above, [Ba2+] = s = 6.22×10^-4 M and [IO3-] = 2s = 1.24×10^-3 M.
iii) The pH of the solution is determined by the hydrolysis of the IO3- ion. The equilibrium expression for this reaction is Ka = [H+][IO3-]/[HIO3]. Since [HIO3] is negligible compared to [IO3-], we can approximate Ka = [H+][IO3-]. Solving for [H+] gives [H+] = Ka/[IO3-] = 1.7×10^-1/1.24×10^-3 = 0.137. The pH is then -log[H+] = -log(0.137) = 0.86.
iv) In a solution of BaCl2, the Ba2+ ion is common to both BaCl2 and Ba(IO3)2. The concentration of Ba2+ is therefore increased by the concentration of BaCl2, which decreases the solubility of Ba(IO3)2. The new solubility is given by s' = (Ksp/4)^(1/3) - [BaCl2] = (1.5×10^-9/4)^(1/3) - 0.01 = 6.22×10^-4 - 0.01 = -0.0094 M. Since solubility cannot be negative, this means that Ba(IO3)2 is completely insoluble in a 0.01 M solution of BaCl2.
v) Since Ba(IO3)2 is insoluble in BaCl2, [Ba2+] = [BaCl2] = 0.01 M and [IO3-] = 0.
vi) Since Ba(IO3)2 is insoluble in BaCl2, there is no IO3- to hydrolyze and the pH is determined solely by the BaCl2. BaCl2 is a strong electrolyte and does not affect the pH, so the pH is 7.
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