Question

For the dissociation reaction of a weak acid in water,

HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq)

the equilibrium constant is the acid-dissociation constant, Ka, and takes the form

Ka=[H3O+][A−][HA]

Weak bases accept a proton from water to give the conjugate acid and OH− ions:

B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)

The equilibrium constant Kb is called the base-dissociation constant and can be found by the formula

Kb=[BH+][OH−][B]

When solving equilibrium-based expression, it is often helpful to keep track of changing concentrations is through what is often called an I.C.E table, where I. stands for Initial Concentration, C. stands for Change, and E. stands for Equilibrium Concentration. To create such a table, write the reaction across the top creating the columns, and the rows I.C.E on the left-hand side.

Initial (M)Change (M)Equilibrium (M)A+ B→AB

Part A

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH. What was the Ka value calculated by the student if the pH of the solution was 2.60?

Express your answer numerically using two significant figures.

Hints

Ka =

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Incorrect; Try Again; 3 attempts remaining

You used the initial concentration of aspirin in the Ka expression. Instead, you need to use the equilibrium concentration. Consider that the amount of aspirin that reacted is equal to the amount of H₃O⁺ produced. How much is left over?

Part B

A 0.100 M solution of ethylamine (C2H5NH2) has a pH of 11.87. Calculate the Kb for ethylamine.

Express your answer numerically using two significant figures.

Hints

Kb =