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

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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?

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

#### Answer:

#### Explanation:

For which:

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

The equation is:

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

We can find the number of moles of HClO:

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

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:

To get

To get this we use

We know that

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

Putting in the numbers:

Now we can use: