Calculate the pH change when water is added to 25.0 ml of 0.250M NaOH to form a 1.00l solution.

The initial [-OH] is 0.250M since NaOH releases 1 mole of -OH per mole of base. Using Kw=[H+][-OH], [H+]=(110^-14)/0.250=410^-14M (note remember units). Hence the initial pH=-log(410^-14)=13.4 (note 3 significant figures since same s.f. as values in question). The diluted [-OH] is 0.250(25/1000)=6.2510^-3M (dilution factor, remember units). Using Kw=[H+][-OH], [H+]=(110^-14)/(6.2510^-3)=1.610^-12M. Therefore the new pH=-log(1.6*10^-12)=11.80. Hence the pH change is 11.80-13.40=-1.60, so a decrease of 1.60pH units. I would write down the question on the whiteboard, and allow the student to work through it, giving guidance and helpful suggestions when they struggled/became stuck - but not simply telling them the answers because it is most beneficial for the student to think through the question for themselves.