Just came across this acids and bases question whilst revising and I've got no idea how to approach it: A 25.0 cm3 sample of 0.0850 mol dm–3 hydrochloric acid was placed in a beaker and 100 cm3 of distilled water were added. Calculate the pH of the new solution formed.
It's only 2 marks so feel like I'm missing something simple - the mark scheme answer is 1.77
I calculated the moles of HCl, so (25/1000)x 0.0850 = 0.002125 but I'm not sure where to go from there. MS isn't particularly helpful as it just states the final answer.
I calculated the moles of HCl, so (25/1000)x 0.0850 = 0.002125 but I'm not sure where to go from there. MS isn't particularly helpful as it just states the final answer.
Okay so you’ve worked out the number of moles of HCl BEFORE dilution.
Now you need to find the conc of HCl after dilution.
For questions that involve dilution of something (so adding water to an acid, adding water to a base etc)
- they will normally give you the concentration and moles of either before or after dilution. - REMEMBER!! the number of moles before and after dilution is always the same
Use this to find the new conc:
number of moles before dilution = number of moles after dilution
so, conc*vol before = conc*vol after
the volume after dilution will be initial volume + volume of water. Remember to convert units!
Not sure why I can’t edit my previous reply but I meant to say (in case anyone else needs help too) questions involving dilutions will always give u the conc and volume of either before or after dilution (NOT conc and no. of moles as you have to work out the no. of moles yourself)