Equal volumes of 0.220 M weak base (Kb=4.0x10^-9) and 0.220 M HCl are mixed. Calculate the pH of the resulting solution

Answer 1

BOH = weak base
BOH + HCl ==> BCl + H2O

So the pH at the equivalence point will be the pH of a solution of the salt, BCl. The B^+ will be hydrolyzed to form
B^+ + HOH ==> BOH + H^+
Ka for B^+ = (Kw/Kb for BOH) - (H^+)(BOH)/(B^+)
(Kw/4.0E-9) = (x)(x)/(0.110-x)
and solve for x = (H^+) and convert to pH.
Where did the 0.110 come from. If you mix equal volumes of equal concns of acid and base you get 1/2 for the concn of the salt or 0.220/2 = 0.110.
Kw is 1E-14 . You probably can avoid a quadratic equation if you assume x will be small and call 0.110-x = 0.110.

Answer 2

Equal volumes of 0.220 M weak base ( 𝐾b=4.0×10−9 ) and 0.220 M HCl are mixed.

Calculate the pH of the resulting solution.

Answer 3

pH = 3.28