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InterviewSolution
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Calculate the pH value of `(i) 10^(-2) "molar" HNO_(3) " solution " (ii) " 0.03 N HCl solution " (iii) 0.001 N H_(2)SO_(4) "solution "`. |
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Answer» (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` |
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