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. `Ca(OH)_(2)=10xx0.2xx2=4 mEq`
`HCl=25xx0.1=2.5 mEq`
Base left `=4-2.5=1.5 mEq`
Total volume `=10+25=35 ml`
`[overset(Θ)(O)H]=1.5/35=0.0428=0.043`
`pOH=-log[0.043]=-log[43-10^(3)]`
`=-[1.633-3]=1.366`
`pH=14-1.366=12.63`
b. `H_(2)SO_(4)=10xx0.01xx2=0.2 mEq`
`Ca(OH)_(2)=10xx0.01xx2=0.2 mEq`
Salt of `S_(A)//S_(B)` is formed.
`pH=7`
c. `H_(2)SO_(4)=10xx0.1xx2=2 mEq`
`KOH=10xx0.1= 1 mEq`
Acid left `=2-1=1 mEq`
Total volume `=10+10=20 ml`
`[H^(o+)]=1/20=0.05`
`pH=- log[0.05]=-log[5-10^(2)]`
`=-[0.69-2]=1.30`