Problem: If Kb for NX3 is 4.5×10−6 , what is the the pKa for the following reaction?HNX3+(aq) + H2O(l) ⇌ NX3(aq) + H3O+(aq)Express your answer numerically to two decimal places.

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We are asked to find pKa for the following reaction

HNX3+(aq) + H2O(l) ⇌ NX3(aq) + H3O+(aq)


    NX3     +     H2O      ⇌      HNX3+     +     OH-
(weak base)         (acid)             (conjugate acid)     (conjugate base) 


Hence, the Ka of HNX3+(aq) can be calculated from the Kb of NX3. Ka and Kb are connected by the autoionization constant of water (Kw) by the following equation:

Kw=Ka·Kb

Kw is a constant and is temperature-dependent. The value of Kw at the given temperature (25°C) is:

Kw = 1.0x10-14             @ 25°C


On the other hand, pKa can be calculated from Ka using the following equation:

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Problem Details

If Kb for NX3 is 4.5×10−6 , what is the the pKa for the following reaction?

HNX3+(aq) + H2O(l) ⇌ NX3(aq) + H3O+(aq)

Express your answer numerically to two decimal places.

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