1.

Calculate the pH of 0.125M of H_(2)SO_(4).

Answer»

SOLUTION :`H_(2)SO_(4)to2H^(+)+SO_(4)^(2-)`
`[H^(+)]=2[H_(2)SO_(4)]=2xx0.125=0.25`
`pH=-log_(10)[H^(+)]=-log_(10)[0.25]=-log_(10)[2.5xx10^(-1)]`
`pH=-log2.5-log10^(-1)=-0.3979+1=0.6021`.


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