Calcuate the `pH` value of (assume `100%` ionization)
(i) `10^(-2)` molar `HNO_(3)` solution
(ii) `0.03M HCl` solution `(log 3=0.4771)`
(iii) `0.0005 M H_(2)SO_(4)` solution

Explanation

(i) `HNO_(3)` completely ionizes as : `HNO_(3)+H_(2)O rarr H_(3)O^(+)+NO_(3)^(-)`
`:. [H_(3)O^(+)]=10^(-2)M` (Given)
`pH=-log[H_(3)O^(+)]=-log(10^(-2))=-(-2 log 10) = 2`
(ii) HCl completely ionizes as : `HCl + H_(2)O rarr H_(3)O^(+) + Cl^(-)`
`:. [H_(3)O^(+)]=[HCl]=0.03 N ` (Given)
`=3xx10^(-2)N = 3xx10^(-2)M ` (HCl is monobasic, Eq. mass = Mol. mass)
`:. pH = - log [H_(2)O^(+)]=- [log 3 xx10^(-2)]=-(log 3 + log 10^(-2))=- (0.4771-2) = 1.5229`
(ii) `H_(2)SO_(4) ` completely ionizes as : `H_(2)SO_(4) + 2H_(2)O rarr 2H_(3)O^(+)+SO_(4)^(2-)`
`{:(,[H_(3)O^(+)]=2xx[H_(2)SO_(4)],,[1 "molecule of" H_(2)SO_(4) "gives"2H_(3)O^(+) "ions"]^(**),,),("But",H_(2)SO_(4)=0.001 N = 0.001 xx49 "g/litre",,("Eq. mass of " H_(2)SO_(4)=49),,),(," "=(0.001xx49)/(98) "moles/litre",,("Mol. mass of " H_(2)SO_(4)=98),,),(," "=0.0005 M ,,("Directly, Molarity"("Normality")/("Basicity")),,):}`
`:. [H_(3)O^(+)]=2xx[H_(2)SO_(4)]=2xx0.0005M = 0.001 M = 10^(-3)M`
`:. pH = - log [H_(3)O^(+)]=-log (10^(-3))=3`