Question

A 25.00 mL sample of 0.300 M HClO(aq), hypochlorous acid, is titrated with 30.00 mL of 0.250 M LiOH. For hypochlorous acid, Ka = 2.910-8

a. b. c.

d.

e. f.

g.

Label each as a strong or weak acid; strong or weak base; acidic, basic or neutral salt:

LiOH ________________ HClO _________________ LiClO ________________ Write the net ionic neutralization reaction for this titration mixture.

Calculate the initial moles of HClO and LiOH and set up a change table for the reaction written in part b. (Show these calculations below or attach work!)

Based on the substances present after neutralization, can the Henderson-Hasselbach equation be used? Explain!

Write the hydrolysis reaction for the hypochlorite ion, ClO-. Show the calculations for the following:

[ClO-] = _______ Kb = ___________ [OH-] = ____________ What is the pH of this titration mixture?

Answer 1

no of moles of HClO = molarity * volume in L
                     = 0.3*0.025 = 0.0075 moles
no of moles of LiOH   = molarity* volume in L
                      = 0.25*0.03 = 0.0075 moles
             HClO(aq)   +   LiOH(aq) -------> LiClO(aq) + H2O(l)
           
    H^+ (aq) +ClO^- (aq)    +   Li^+ (aq) +OH^- (aq) -------> Li^+(aq) + ClO^-(aq) + H2O(l)
      H^+ (aq) + OH^-(aq) ----------> H2O net ionic equation

          ClO^- + H2O ----------> HClO + OH^-

     I     0.0075                  0     0
     C       -x                    +x    +x
     E     0.0075-x                 +x    +x
     
   Kb = Kw/Ka
      = 1*10^-14/2.9*10^-8
      = 3.45*10^-7
    kb = [HClO][OH^-]/[ClO^-]
    3.45*10^-7 = x*x/0.0075-x
    3.45*10^-7*(0.0075-x) = x^2
         x = 5*10^-5
     [OH^-] = 5*10^-5M
      POh = -log[OH^-
           = -log5*10^-5
            = 4.3010
       PH   = 14-POH
            = 14-4.3010   = 9.699