Considere uma solução 50 μmol ⋅ L−1\pu{50 \mu mol.L-1}50 μmol⋅L−1 de hidróxido de potássio, KOH\ce{KOH}KOH.
Assinale a alternativa que mais se aproxima do pH da solução em 25 °C\pu{25 \degree C}25 °C.
[OHX−]≈[KOH]=5⋅10−5 mol ⋅ L−1[\ce{OH^-}]\approx[\ce{KOH}]=\pu{5e-5 mol.L-1}[OHX−]≈[KOH]=5⋅10−5 mol⋅L−1 [HX3OX+][OHX−]=10−14 mol2 L−2 ⟹ 5⋅10−5 mol ⋅ L−1[HX3OX+]=10−14 mol2 L−2 ⟹ [\ce{H3O^+}][\ce{OH^-}]=\pu{10^{-14} mol^2 L^{-2}}\implies\pu{5e-5 mol.L-1}[\ce{H3O^+}]=\pu{10^{-14} mol^2 L^{-2}}\implies[HX3OX+][OHX−]=10−14 mol2L−2⟹5⋅10−5 mol⋅L−1[HX3OX+]=10−14 mol2L−2⟹ ⟹ [HX3OX+]=2⋅10−10 mol ⋅ L−1 ⟹ pH=−log(2e−10)≈9,7\implies [\ce{H3O^+}]=\pu{2e-10 mol.L-1}\implies pH=-log(2e-10)\approx 9,7 ⟹[HX3OX+]=2⋅10−10 mol⋅L−1⟹pH=−log(2e−10)≈9,7