[Feron]

Hi, i didn't know this forum existed until i was browsing google some 20 minutes ago - and i love chemstry and will definitley hang around here more!! 

Im doing an assignment and its a titration between 250cm3 calcium hydroxide with a concentration of 0.05398 mol/dm3 and Hydrochloric acid.
the acid is 2.00 mol/dm3 which is obviously way to strong to react with the calcium hydroxide.  I've left my textbook at school and i would really appreciate help on how much to dilute the hydrochloric acid to make it the same concentration as the calcium hydroxide.

Thanks in advance
- Feron

[Yggdrasil]

You can use the formula C₁V₁ = C₂V₂ to calculate dilutions, where C₁ is the concentration of your stock solution, V₁ is the volume of the stock solution you use, C₂ is the concentration of the diluted solution, and V₂ is the volume of your dilution (volume of stock solution + volume of water added).

[Feron]

so 2mol/d3 x 0.25 = 0.05398 x unknown volume of (water + 0.25)

= 0.5 / 0.05398 = 9.263

therefore water added  = 9.263 - 0.25 = 9.13dm3 to 0.25dm3 of HCl

im not sure thats entirely correct??

[Yggdrasil]

Your calculations are correct, but the logic behind setting them up is not so much correct.

Instead of diluting 250mL of HCl to make 9L of diluted HCl, why not choose a final volume that seems reasonable, then figure out how much of your concentrated HCl you need to make your dilution?

[Feron]

Yeah that was just an example.

i set V2 to 0.25dm3 and worked out how much acid i would need.

thanks!!