A total of 100. mL of 0.1000M HClO (Ka= 3.0x10-8)is titrated with 0.0100M KOH at 25C?

A: Calculate initial pH of the HClO solution

B: What is the pH of the solution at the equivalence point, and how many
mL of KOH solution must be added to reach the equivalence point?

1 Answer
Michael
Nov 17, 2016

#sf((A))#

#sf(pH=4.26)#

#sf((B))#

#sf(pH=9.74)#

#sf(1000color(white)(x)ml)# must be added.

Explanation:

#sf((A))#

#sf(HClO)# is a weak acid and dissociates:

#sf(HClOrightleftharpoonsH^(+)+OCl^(-))#

For which:

#sf(K_a=([H^+][Cl^-])/([HClO])=3.0xx10^(-8)color(white)(x)"mol/l")#

Since we have a weak acid we can rearrange and take -ve logs of both sides to get:

#sf(pH=1/2[pK_a-loga])#

#sf(a)# is the concentration of the acid which we can assume to be very close to the equilibrium concentration.

#sf(pK_a=-log(K_a)=-log(3.0xx10^(-8))=7.522)#

#:.##sf(pH=1/2[7.522-(-1)]=4.26)#

#sf((B))#

The equation is:

#sf(HClO+KOHrarrKClO+H_2O)#

This tells us that 1 mole of HClO is equivalent to 1 mole of KOH.

We can find the number of moles of HClO:

#sf(n=cxxv=0.1000xx100/1000=0.01)#

The equation tells us that the no. moles of KOH must be the same:

#sf(nKOH=0.01)#

#sf(c=n/v)#

#:.##sf(v=c/n=0.01/0.01=1"L")#

This is the volume of KOH solution required to reach the equivalence point.

This is a badly set question as you would never set up a titration that required such a large volume. The acid is far too concentrated.

It should be diluted 10 times and only 25 ml used. That would give a sensible end - point.

We have added 0.01 moles of HClO to 0.01 moles of KOH to make 0.01 moles KClO.

We can use a corresponding expression to get the pH of a weak base:

#sf(pOH=1/2[pK_b-logb])#

To get #sf(pK_b)# we use:

#sf(pK_a+pK_b=14)#

#:.##sf(pK_b=14-pK_a=14-7.522=6.478)#

#sf(b=[ClO^-])#

To get this we use #sf(c=n/v)#

We know that #sf(n=0.01)#

The new total volume = 1000ml + 100ml = 1100 ml = 1.1 L

#:.##sf(c=[ClO^-]=0.01/1.1=0.00909color(white)(x)"mol/l")#

Putting in the numbers:

#sf(pOH=1/2[6.478-(-2.041)]=4.2595)#

Now we can use:

#sf(pH+pOH=14)#

#:.##sf(pH=14-pOH=14-4.2595=9.74)#

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