Map A General Chemistry University Science Books 4th Edition presented by Saplin

Question

Map A General Chemistry University Science Books 4th Edition presented by Sapling Learning Donald McQuarrie. Peter A. Rock Ethan Gallogly Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.100 M HCIO(aq) with 0.100 M KOH(aq). The ionization constant for HCIO can be found here. Number (a) before addition of any KoH Number (b) after addition of 25.0 mL of KOH Number (c) after addition of 40.0 mL of KOH Number (d) after addition of 50.0 mL of KOH Number (e) after addition of 60.0 mL of KOH Previous Check Answer Next Exit Hint

Explanation / Answer

a) Before addition of any KOH

The ionisation eqaution for HClO is

HClO --> H+ + ClO-     Ka = 4 x 10-8

Let us assume [H+] =[ClO-] = X, then

Ka = [H+][ClO-]/[Acid] = X^2/0.1 M = 4 x 10-8

X = 6.324 x 10-5 M = [H+]

pH = -log[H+] = -log [ 6.324 x 10-5] = 4.2

b) After addition of 25 ml KOH

    HClO + KOH = KClO + H2O

In the above equation HClO and KOH were reacted in 1: 1 ratio to produce 1 moles of KClO

Let us calculate the unreacted acid as below

No. of moles of acid = 0.1 M x 0.05 L = 0.005 moles

No. of moles of KOH = 0.1 M x 0.025 L = 0.0025 moles

Hence the remaining Acid = 0.005 - 0.0025 = 0.0025 moles

Concentration of salt and acid in final volume = 50+25 =75 ml = 0.0025 moles / 0.075 L = 0.033 M

According henderson hasselbalch equation pH = pKa +log[salt/acid]

pH = -log[4 x 10-8] + log[0.033/0.033]

      = 7.4

c) After addition of 40 ml KOH

Calculate as above , we will get

Unreacted acid = 0.001 mole in 90 ml, salt = 0.004 moles in 90 ml

pH = -log[4 x 10-8] + log[0.044/0.111]

     = 7

d) After addition of 50 ml KOH

   In this case acid completely reacts to form 0.005 moles of salt

   Concentration of salt = 0.005moles /0.1 L = 0.05 M

   KClO --> K+ + ClO-

   ClO- + H2O --> HClO + OH-

   Kb = [ClO-][OH-]/KClo = 2.5 x 10-7    (Calculate from Ka value)

   Assume OH- = X , then 2.5 x 10-7   = X^2/0.05

   X = 1.118 x 10-4 =[OH-]

   Hence [H+] = 8.94 x 10-11 M

   pH = 10.05

e) after 60 ml of 0.1M KOH

    Apply M1V1 = M2V2 fomula

     M1 = 0.1 x 60 / 110 = 0.054 = OH-

    H+ = 1.8 x 10-13

     pH = 12.7

   

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