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`.

Answer

a) `12.6` , b) `7.00` , c) `1.3`

Explanation

(a) Moles of `H_(3)O^(+)=(25xx0.1)/(1000)=.0025` mol
Moles of `OH^(-)=(10xx0.2xx2)/(1000)=.0040` mol
Thus, excess of `OH^(-)` = .0015 mol
`[OH^(-)]=(.0015)/(35xx10^(-3))"mol"//"L"` = .0428
pOH = `-log[OH]` = 1.36
pH = 14 - 1.36 = 12.63 (not matched)
(b) Moles of `H_(3)O^(+)=(2xx10xx0.1)/(1000)=.0002` mol
Moles of `OH^(-)=(2xx10xx0.1)/(1000)=.0002` mol
Since there is neither an excess of `H_(3)O^(+)orOH^(-)`, the solution is neutral. Hence, pH = 7
(c) Moles of `H_(3)O^(+)=(2xx10xx0.1)/(1000)=.002`
Moles of `OH^(-)=(10xx0.1)/(1000)=0.001` mol
Excess of `H_(3)O^(+)` = .001 mol
Thus, `[H_(3)O^(+)]=(.001)/(20xx10^(-3))=(10^(-3))/(20xx10^(-3))=.05`
`thereforepH =-log(0.05)`
= 1.30