O pKa\mathsf{p}K_\mathsf{a}pKa do ácido iodíco é 0,77\pu{0,77}0,77.
Assinale a alternativa que mais se aproxima do pKb\mathrm{p}K_\mathrm{b}pKb do íon IOX3X−\ce{IO3^-}IOX3X−.
HIOX3X (aq)+HX2OX (l)⇌IOX3X−X (aq)+HX3OX+X (aq)\ce{HIO3_{(aq)} + H2O_{(l)} <=> IO3^-_{(aq)} + H3O^+_{(aq)}}HIOX3X(aq)+HX2OX(l)IOX3X−X(aq)+HX3OX+X(aq) Ka=[HX3OX+][IOX3X−][HIOX3]K_a=\frac{[\ce{H3O^+}][\ce{IO3^-}]}{[\ce{HIO3}]}Ka=[HIOX3][HX3OX+][IOX3X−] IOX3X−X (aq)+HX2OX (l)⇌HIOX3X (aq)+OHX−X (aq)\ce{IO3^-_{(aq)} + H2O_{(l)} <=> HIO3_{(aq)} + OH^-_{(aq)}}IOX3X−X(aq)+HX2OX(l)HIOX3X(aq)+OHX−X(aq) Kb=[OHX−][HIOX3][IOX3X−]=[HX3OX+][OHX−]Ka=KwKa ⟹ K_b=\frac{[\ce{OH^-}][\ce{HIO3}]}{[\ce{IO3^-}]}=\frac{[\ce{H3O^+}][\ce{OH^-}]}{K_a}=\frac{K_w}{K_a}\impliesKb=[IOX3X−][OHX−][HIOX3]=Ka[HX3OX+][OHX−]=KaKw⟹ pKb=pKw−pKa=14−0,77=13,23pK_b=pK_w-pK_a=14-0,77=13,23pKb=pKw−pKa=14−0,77=13,23