O pKb\mathsf{p}K_\mathsf{b} da piridina é 8,8\pu{8,8}.

Assinale a alternativa que mais se aproxima do pKa\mathrm{p}K_\mathrm{a} do cátion CX5HX5NHX+\ce{C5H5NH^+}.

Gabarito

CX5HX5NX(aq)+HX2OX(l)CX5HX5NHX+X(aq)+OHXX(aq)\ce{C5H5N_{(aq)} + H2O_{(l)} <=> C5H5NH^+_{(aq)} + OH^-_{(aq)}} Kb=[OHX][CX5HX5NHX+][CX5HX5N]K_b=\frac{[\ce{OH^-}][\ce{C5H5NH^+}]}{[\ce{C5H5N}]} CX5HX5NHX+X(aq)+HX2OX(l)CX5HX5NX(aq)+HX3OX+X(aq)\ce{C5H5NH^+_{(aq)} + H2O_{(l)} <=> C5H5N_{(aq)} + H3O^+_{(aq)}} Ka=[HX3OX+][CX5HX5N][CX5HX5NHX+]=[HX3OX+][OHX]Kb=KwKb    K_a=\frac{[\ce{H3O^+}][\ce{C5H5N}]}{[\ce{C5H5NH^+}]}=\frac{[\ce{H3O^+}][\ce{OH^-}]}{K_b}=\frac{K_w}{K_b}\implies pKa=pKwpKb=148,8=5,2pK_a=pK_w-pK_b=14-8,8=5,2