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)10 mL of 0.2 M `Ca(OH)_2=(10xx0.2)/1000`
`=2xx10^(-3)` moles
25 mL of 0.1 M HCl =`(25xx0.1)/1000`
`=2.5xx10^(-3)` moles
`Ca(OH)_2+2HCl to CaCl_2 +2H_2O`
1 mole of `Ca(OH)_2` reacts with 2 moles of HCl `2.5xx10^(-3)` moles of HCl will react with `1.25xx10^(-3)` mole of `Ca(OH)_2` .
`therefore Ca(OH)_2` left =`2xx10^(-3)-1.25xx10^(-3)`
`=0.75 xx10^(-3)` mole
Total volume =10+25=35 mL
Molarity of `Ca(OH)_2=(0.75xx10^(-3)xx1000)/35`
=0.0214 M
`[OH^-]=2xx0.0214=4.28xx10^(-2)` M
pOH=-log `(4.28xx10^(-2))=-(-2+0.6314)`
1.368
pH =14- 1.368 =12.632
(b) 10 mL of 0.01 `H_2SO_4 =(10xx0.01)/1000`
`=0.1xx10^(-3)` mole
10 mL of 0.01 M `Ca(OH)_2 =(10xx0.01)/1000`
`=0.1xx10^(-3)` mole
`Ca(OH)_2 +H_2SO_4 to CaSO_4 + 2H_2O`
1 mole of `Ca(OH)_2` reacts with 1 mole of `H_2SO_4`
`therefore 0.1 xx10^(-3)` moles of `Ca(OH)_2` will react completely with `0.1xx10^(-3)` mole of `H_2SO_4`.
The resulting solution will be neutral with pH=7.0
(c) 10 mL of 0.1 M `H_2SO_4 =(10xx0.1)/1000`
`=1xx10^(-3)` mole
`2KOH +H_2SO_4 to K_2SO_4+ 2H_2O`
1 mole of KOH reacts with 0.5 mole of `H_2SO_4`
`therefore 1xx10^(-3)` mole of KOH will react with `0.5xx10^(-3)` mole of `H_2SO_4` .
`H_2SO_4` left =`(1-0.5)xx10^(-3)=0.5xx10^(-3)`
Volume of solution =10+10=20 mL
Molarity of solution =`(0.5xx10^(-3) xx1000)/20`
`=2.5xx10^(-2)`M
`therefore [H^+]=2xx2.5xx10^(-2) M =5.0xx10^(-2)` M
pH=-log `(5.0xx10^(-2))`=-(-2+0.699)
=1.3