(Solved): urgently please Consider the titration of 30.0mL of 0.250M weak base ClO(Kb=3.610 ...

Expert Answer

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(1) Firstly we need to consider the basicity of the weak base ClO- for determining the pH of the solution before any acid has been added. Since ClO- is a weak base, it will undergo partial ionization in water to form ClO- and hydroxide ions. The equilibrium expression for the ionization reaction is as follows:
ClO- + H2O ? ClO- + OH-
The Kb expression for ClO- is:
Kb = [ClO-][OH-]/[ClO-]
Since we know the Kb value for ClO- is 3.6×10-7, we can use this expression to calculate the hydroxide ion concentration [OH-]:
3.6×10-7 = [ClO-][OH-]/[ClO-]


The pOH of the solution is:
pOH = -log[OH^-] = -log(1.35×10^?4) = 3.87
Since pH + pOH = 14, the pH of the solution before any acid has been added is:
pH = 14 - pOH = 14 - 3.87 = 10.13
Therefore, the pH of the solution before any acid has been added is 10.13.

(2) When half of the weak base has been neutralized, we have reached the halfway point of the titration, or the half-equivalence point. At this point, half of the original amount of ClO- has been converted to Cl- and the other half remains as ClO-. Since the reaction is a one-to-one reaction, the moles of HBr added is equal to the moles of ClO- that have been neutralized. Therefore, at the half-equivalence point, the concentration of ClO- is half of its original concentration, or 0.125 M. Now with the help of the Kb expression, we can calculate the [OH-] concentration:
Kb = [ClO-][OH-]/[ClO-] [OH-] = Kb*[ClO-]/[ClO-] = Kb = 3.6×10-7
The pOH of the solution is:
pOH = -log[OH-] = -log(3.6×10-7) = 6.44
Since pH + pOH = 14, the pH of the solution at the half-equivalence point is:
pH = 14 - pOH = 14 - 6.44 = 7.56
Therefore, the pH of the solution when half of the weak base has been neutralized is 7.56.

(3) The net ionic balanced equation for the titration is:
  

Please refer to the solution for detailed explanation.