Calculate pH of
a. `10^(-3)M HCl`
b. `10^(-3) MH_(2) HO_(4)`
c. `10^(-6) M HNO_(3)`
d. `0.02 MH_(2)SO_(4)`

Explanation

a. `pH=-log[H^(+)]=-log[1xx10^(-3)]=3`
b. `H_(2)SO_(4)to2H^(+)+SO_(4)^(2-)`
Since 1 mole of `H_(2)SO_(4)` gives 2 moles of `H^(+)` Concentration of `H^(+)=10^(-3)xx2=2xx10^(-3)`
`pH=-log[2xx10^(-3)]=2.699`
c. `10^(-6)M HNO_(3)`
`pH=-log[10^(-6)]=6`
d. `0.02 M H_(2)SO_(4)`
`[H^(+)]=0.04`
( `:'` mole of `H_(2)SO_(4)` gives 2 moles of `H^(+)`)
`pH=-log[4xx10^(-2)]=1.4`