Calculate the pH of the resultant mixtures :
10 mL 0.2 M `Ca(OH)_2`+ 25 mL 0.1 M HCl

Explanation

Molarity =Mol/Litre, So, Mol=Molarity x Litre
10 mL 0.2 M `Ca(OH)_2=10/1000` L x 0.2 M
`=2/1000 "mol" Ca(OH)_2`
=2 millimole `Ca(OH)_2`
25 mL 0.1 M HCl = `25/1000` L x 0.1 M
`=2.5/1000`mol HCl
=2.5 millimole HCl
Remaining substrate of neutralization:
`{:(Ca(OH)_2+, 2HCl to , CaCl_2+,2H_2O),("1 mol","2 mol","1 mol","2 mol"):}`
Here, 2 mol HCl react with 1 mol `Ca(OH)_2`
`therefore 2.5/1000` mol HCl `to 1.25/1000` mol `Ca(OH)_2` react.
`therefore` The remaining mol of `Ca(OH)_2` after neutralisation.
`(("Initial"),(2.0/1000))-(("After reaction"),(1.25/1000))`
`=0.75/1000` mol `Ca(OH)_2` is remainly
Molarity of remaining `Ca(OH)_2=("mol"/"Total volume (L)")`
`=(0.75/1000)1000/35`=0.02143 M
`[OH^-]=2[Ca(OH)_2]=2xx0.02143`
=0.04286 M =`4.286xx10^(-2)` M
pOH=-log `[OH^-]=-log (4.286xx10^(-2))`
=-(-1.3678)=1.3679
pH=(14.0-pOH)=(14.0-1.3679)
=12.632 =12.63