[Invisible]

You have a solution of approximately 0.196 molL-1 NaOH, but you need to confirm its concentration by titration.  You have a solution of 3.92 molL-1 HCl.  The HCl is too concentrated, you will need to dilute it.
Good practice says that you should have approximately 20mL of HCl run from the burette to the 20mL of NaOH in the flask.  Show by calculation and reason the dilution of HCl that you will need to achieve to be carrying out good practice.

[AWK]

Show your attempt first

[Invisible]

Ok well this is probably all wrong but I have this formula in my book to work out dilutions.
new concentration=old concentration * (volume taken/flask volume)
so ? = 3.92 molL-1 * (?/0.02L?)
Now i'm stuck...

[Borek]

Wrong end.

You need to calculate how much acid you will need to titrate 20 mL of your NaOH solution, calculate what concentration of acid is necessary if that amount of acid should be present in 20 mL of titrant, then - finally - think about ways of diluting.

These pages may come handy:

http://www.titrations.info/titration-calculation

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

[Invisible]

thanks but I still cant work it out.
first I worked out the moles of NaOH in 20mL.  n=c/v  =0.196molL-1*0.02L  =0.00392mol.
so the equation is NaOH + HCl --> NaCl +H2O.  the mole ratio of NaOH to HCl is 1:1 so the moles of HCl is 0.00392mol.  am i on the right track here??
next i wasn't sure but calculated the concentration of HCl by c= 0.00392/0.02  =0.196molL-1.
Not quite sure where to go with this...

[Borek]

Not surprisingly to titrate 20 mL of 0.196 M NaOH you need 20 mL of 0.196M HCl. As they react 1:1 this is rather expected.

Now you have to think of way of diluting your HCl to about 0.2M (it doesn't have to be exactly 0.196). How much titrant solution do you want to prepare?

[Invisible]

I don't really understand the next step.
If I go v(HCl)=n/c   v=0.196mol / 3.92molL-1    v=0.05L 
So is the dilution of HCl 50mL?

[Invisible]

Is it right??