Question

# 3. A 25.00mL sample of 0.300 M HClO(aq), hypochlorous acid, is titrated with 30.00mL of 0.250M...

3. A 25.00mL sample of 0.300 M HClO(aq), hypochlorous acid, is titrated with 30.00mL of 0.250M LiOH. For hypochlorous acid, Ka= 2.9 x10-8

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

b.Write the net ionic neutralization reaction for this titration mixture.

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

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

e.Write the hydrolysisreaction for the hypochloriteion, ClO-.

f.Show the calculations for the following:[ClO-] = _____ Kb= _____ [OH-] = _______

g.What is the pH of this titration mixture?

a)

HClO = weak acid, since forms only partial ionization

LiOH = strong base, donates 100% OH-

LiClO = conjugate base ClO- acts as a base

b)

net ionic neutralization

HClO(aq) + OH-(aq) -- >H2O(l) + ClO-(aq)

c)

mol of HClO = MV = (0.3)(25*10^-3) = 0.0075

mol of LiOH = MV = 0.25*30*10^-3 = 0.0075

HClO = 0.0075 - x

ClO- = 0 + x

d)

this can't be used

since there is only ClO- left , no HClO presence

so there will be hydrolysis only, no buffer

e)

ClO-(aq) + H2O(l) <-> HClO(aq) + OH-(aq)

f)

Kb = Kw/Ka = (10^-14)/(2.9*10^-8) = 3.4482*10^-7

Kb = [HClO][OH-]/[ClO-]

mmol of ClO- = 0.3*25 / (25+30) =

(3.4482*10^-7) = x*x/(0.1363-x)

x^2 + (3.4482*10^-7) x + (0.1363)((3.4482*10^-7) ) = 0

x = 0.000216

[HClO]= [OH-] = 0.000216

[ClO-] = 0.1363-0.000216 = 0.136084

pOH = -log(0.000216) = 3.665

pH = 14-poH = 14-3.665

pH = 10.335

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