A monoprotic acid, HA, is dissolved in water: The equilibrium concentrations of the reactants and products are [HA] = 0.160 M [H] = 4.00

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William Appel · Accepted Answer

A monoprotic acid dissociates according to the reaction equation:  HA(aq)------&amp;gt; H+(aq) + A-(aq)  The equilibrium concentrations in M satisfy the relation:  Ka = [H+]*[A-]/[HA]  For this acid  Ka=4.00×10−4⋅4.00×10−40.16M  Ka=10−6  Hence,  pKa=−log⁡10(Ka)=−log⁡10(10−6)=6

reinosodairyshm · Accepted Answer

A monoprotic acid dissociates according to the reaction equation:  HA(aq) -------&amp;gt; H+(aq) + A-(aq)  The equilibrium concentrations in M satisfy the relation:  Ka = [H+]·[A-]/[HA]  For this acid  Ka=2.00×10−4·2.00×10−40.26M  =1.54×10−7  Hence,  pKa=−log⁡10(Ka)=−log⁡10(1.54×10−7)=6.81

A monoprotic acid, HA, is dissolved in water: The equilibrium

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