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

Explanation

(a)meq of `Ca(OH)_(2)=2xx0.2xx10=4`
meq of `HCl=1.01xx25=2.5`
`implies`Solution is basic. Excess meq of base = 1.5 `implies[OH^(-)=1.5/35=0.043M impliespH = 14 + log[OH^(-)] = 12.63`
Note : Molarity of `OH^(-)` ions = Normality of `OH^(-)` ions as n-factor of `OH^(-)` ions is 1.
(b)Observe clearly that :meq of `H_(2)SO_(4) =` meq of `Ca(OH)_(2)`
So it is neutral solution as salt is of strong Acid and Strong base. pH = 7
(c) meq of acid = `2 xx 0.1 xx 10 = 2`
meq of base = `1 xx 0.1 xx 10 = 1implies`Solution is acid.
`[H^(+)]_(excess)=1/20 implies pH=-log 1/20=1.3`
Note that molority of `H^(+)` ions is some as its normally as n-factor of `H^(+)`