[jiunshan]

I have done a titration lab....but I need help in finding the number of moles of potassium hydroxide in the original undiluted solution.

KOH + HCl --> KCl + H₂O

Volume of KOH used: 40.0cm³ (undiluted)

40.0cm³KOH diluted with distilled water to 250cm³


25cm³ of the KOH solution required 19.3cm³of the HCl for complete reaction.

The concentration og the HCl used is 0.1gdm^-3 AM I RIGHT??

Then...

1. Number of moles of KOH in 1dm³ of diluted solution = 7.72 * 10^-3 mol.  AM I RIGHT?

2. What is the number of moles of KOH in the original undiluted solution??

How to solve this problem?

TQ

[Borek]

The concentration og the HCl used is 0.1gdm^-3 AM I RIGHT??

No way to tell, but either it was given and you just forgot to write it down, or HCl standarization was a part of the lab (and you missed that step). First option is much more likely.

[jiunshan]

The concentration og the HCl used is 0.1gdm^-3 AM I RIGHT??

No way to tell, but either it was given and you just forgot to write it down, or HCl standarization was a part of the lab (and you missed that step). First option is much more likely.

'coz it is using standard hydrochloric acid. So the molarity for standard HCl is 0.1M?

[Borek]

There is no such thing as "standard hydrochloric acid". It doesn't rule out fact that concentration could be 0.1M.

In lab practice before you use HCl solutions you should prepare them so that they are about 0.1M (or 0.01M or something, depends on what you are going to do) and then you standardize such solution to know its exact concentration. It can be 0.1234M or 0.09876M or anything, you just have to know exact concentration with four significant digits.

It is possible to prepare 0.1000M solution, but it requires time and skill and is a waste of time IMHO. All it does it makes calculations easier, but if you use calculator it doesn't mean anything.