Explanation Given: The pH of the solution is 9 .29 . The given solution of NaA is assumed to be completely dissociated into Na + and A − . Na + is a group I cation and does not react with water therefore do not alter the value of pH . The A − ion is a weak base and it reacts with water to give HA . The reaction of A − with water is water to give HA is, A − ( a q ) + H 2 O ( a q ) ⇌ HA ( a q ) + OH − ( a q ) Formula used: At equilibrium, the equilibrium constant expression is expressed by the formula, K a = Concentration of products Concentration of reactants Where, K a is the acid dissociation. The relationship between equilibrium constant of conjugate acid and conjugate base is, K w = K a × K b Where, K b is the ionization constant. K a is the dissociation constant. K w is the ionic product of water Calculation: The equilibrium constant expression for the above reaction is, K b = [ OH − ] [ HA ] [ A − ] ....(1) The pH of the solution is 9 .29 . The value of pOH is calculated by the formula, pOH = 14 − pH =14 − 9 .29 =4 .71 The value of pOH is calculated by the formula, pOH = − log [ OH − ] Rearrange the above expression to find the value of [ OH ] . [ OH − ] = 10 − pOH Substitute the value of pOH in the above expression, [ OH − ] = 10 − pOH [ OH − ] = 10 − 4

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Chapter 2, Problem 86E

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To determine: The value of Ka for the parent acid HA .

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Chapter 2, Problem 85E

Chapter 2, Problem 87AE

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The value of K a for the parent acid HA .

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To determine: The value of Ka for the parent acid HA .

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Chapter 2 Solutions