Problem: If enough of a monoprotic acid is dissolved in water to produce a 0.0150 M solution with a pH of 6.05, what is the equilibrium constant for the acid?

FREE Expert Solution

The dissociation of HA is as follows:

HA(aq) + H2O(l)  H3O+(aq) + A(aq)


From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.



The Ka expression for HA is:

Ka=productsreactants=[H3O+][A-][HA]


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If enough of a monoprotic acid is dissolved in water to produce a 0.0150 M solution with a pH of 6.05, what is the equilibrium constant for the acid?

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