Calculate the pH of the resultant mixture:
a. `10 mL` of `0.2M 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) pH of 10 mL of 0.2 M `Ca (OH)_(2)` and 25 mL of 0.1 M HCI solution
Molarity of `Ca(OH)_(2)` solution on mixing `=((0.2 M ) xx (10mL))/((35 mL)) = 0.057 M`
`[OH^(-)]` in solution `=2xx 0.057 M = 0.114 M`
Molarity of HCI solution on mixing `=((0.1 M) xx (25 ML))/((35 mL)) =0.071 M`
`[H^(+)]` in solution `=0.071 M`
`[OH^(-)]` in solution after neutralisation `=(0.114 - 0.071) = 0.043 M`
`pOH =- log [OH^(-)] =- log (4.3 xx 10^(-2))`
`=- (log 4.3 -2 log 10) =(2 - log 4.3)`
`=(2- 0.633) = 1.367`
`pH =14 -pOH =14 - 1.367 = 12.63`
(b) pH of 10 mL of 0.01 M `H_(2)SO_(4)` and 10 mL of 0.01 M `Ca(OH)_(2)` solutions
Molarity of `H_(2)SO_(4)` solution on mixing `=((0.01 M) xx (10 mL))/((20 mL)) =0.005 M`
`[H^(+)]` in solution `=0.005 xx 2= 0.01 M`
Molarity of `Ca(OH)_(2)` solution on mixing `=((0.01 M) xx (10 mL))/(20 mL)) = 0.005 M` ,
`[OH^(-)]` in solution `=0.005 xx 2 = 0.01 M`
Since `[H^(+)]` and `[OH^(-)]` in the solution are the same , it is neutral in nature.
pH of solution =7
(c) 10mL of 0.1 m `H_(2)SO_(4)` and 10 mL of 0.01 M KOH solutions
Molarity of `H_(2)SO_(4)` solution on mixing `=((0.1 M) xx (10 mL))/((20 mL)) =0.05 M`
`[H^(+)]` in solution `=0.05 xx 2 =0.1 M`
Molarity of KOH solution on mixing `=((0.1 M) xx (10 mL))/((20 mL)) =0.05 M`
`[OH^(-)]` in solution `=0.05 M`
`[H^(+)]` in solution after neutralisation `=0.1 0.05 =0.05 m`
`pH =- log [H^(+)] =- log (5xx 10^(-2))`
`=- (log 5-2 log 10) =(2 - log 5)`
`=2 -0.6989 = 1.301`