Question

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?

Answer 1

`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)`