Calculate the pH of the resultant mixtures :
(a) 10 mL of 0.2 M `Ca(OH)_(2)` + 25 mL of 0.1 M HCl
(b)10 mL of 0.01 M `H_(2) SO_(4)` + 10 mL of 0.01 M `Ca(OH_(2))`
(c) 10 mL of 0.1 M `H_(2) SO_(4)` + 10 mL of 0.1 M KOH

Explanation

(a) mL of 0.2 M `Ca(OH)_(2)=10xx0.2` millimoles = 2 millimoles of `Ca(OH)_(2)`

25mL of 0.1 M HCl `= 25xx0.1` millimoles = 2.5 millimoles of HCl
`Ca(OH)_(2)+2HCl rarr CaCl_(2)+2 H_(2)O`
1 millimole of `Ca(OH)_(2) + 2 HCl rarr CaCl_(2) + 2H_(2)O`
1 millimole of `Ca(OH)_(2)` reacts with 2 millimoles of HCl
`:.` 2.5 millimoles of HCl will react with 1.25 millimoles of `Ca(OH)_(2)`
`:. Ca(OH)_(2) ` left `= 2-1.25 = 0.75` millimoles (HCl is the limiting reactant)
Total volume of the solution = 10 + 25 mL = 35 mL
`:.` Molarity of `Ca(OH)_(2)` in the mixture solution `=(0.75)/(35) M = 0.0214M`
`:. [OH^(-)]=2xx0.0214 M = 0.0428 M = 4.28 xx 10^(-2)`
`pOH = - log (4.28xx10^(-2))=2-0.6314=1.39=686~=1.37`
`:. pH = 14 - 1.37 = 12.63`
(b) 10 mL of 0.01 M `H_(2)SO_(4) = 10xx0.01` millimole = 0.1 millimole
10 mL of 0.01 M `Ca(OH)_(2) = 10 xx 0.01` millimole = 0.1 millimole
`Ca(OH)_(2) + H_(2)SO_(4) rarr CaSO_(4) + 2H_(2)O`
1 mole of `Ca(OH)_(2) ` reacts with 1 mole of `H_(2)SO_(4)` ltbvrgt `:. ` 0.1 millimole of `Ca(OH)_(2)` will react completely with 0.1 millimole of `H_(2)SO_(4)`. Hence, solution will be neutral with pH = 7.0
(c) 10 mL of 0.1 M `H_(2)SO_(4) = 1 ` millimole
10 mL of 0.1 M KOH = 1 millimole
`2KOH + H_(2)SO_(4) rarr K_(2)SO_(4) + 2 H_(2)O`
1 millimole of KOH will react with 0.5 millimole of `H_(2)SO_(4)`
`:. H_(2)SO_(4) ` left = 1 - 0.5 = 0.5 millimole
Volume of reaction mixture = 10 + 10 = 20 mL
`:. ` Molarity of `H_(2)SO_(4)` in the mixture solution `=(0.5)/(20) = 2.5xx106(-12) M`
`[H^(+)]=2xx(2.5xx10^(-2))=5xx10^(-2)`
`pH = - log (5xx10^(-2))=2-0.699=1.3`