PH at the equivalence point?

PH at equivalence point? Chem help?

• Answer:

  At the equivalence point, you have a solution of BH+ in water. This acts as a weak acid by the reaction: BH+ <--> B + H+ Ka = [H+][B]/[BH+] For any acid-base pair, Ka X Kb = Kw = 1 X 10^-14 So, for this conjugate acid, Ka = 1X10^-14 / 8.4 X 10^-7 = 1.19 X 10^-8 Now, you need to calculate the concentration of HB+ in the solution: Moles B initially = 0.010 L X 0.33 mol/L = 3.3X10^-3 mol B Volume of HCl added = 3.3X10^-3 mol / 0.200 mol/L = 0.0165 L = 16.5 mL Concentration of BH+ = 3.3X10^-3 mol / 0.0265 L = 0.1245 M Let x = [H+] = [B], and assume x will be small compared to 0.1245. Then, Ka = 1.19 X 10^-8 = x^2 / 0.1245 x = 3.84 X 10^-5 = [H+] pH = - log 3.84X10^-5 = 4.42