Consider the following curve (Figure 1) for the titration of a weak monoprotic a

Question

Consider the following curve (Figure 1) for the titration of a weak monoprotic acid with a strong base and answer each of the following questions.

Part A

What is the pH at the equivalence point?

Express your answer as whole number.

Part B

What is the volume of added base at the equivalence point?

Express your answer using one significant figure.

Part C

At what volume of added base is the pH calculated by working an equilibrium problem based on the initial concentration and Ka of the weak acid?

Express your answer using one significant figure.

Part D

At what volume of added base does pH=pKa?

Express your answer using two significant figures.

Part E

At what volume of added base is the pH calculated by working an equilibrium problem based on the concentration and Kb of the conjugate base?

Express your answer using one significant figure.

Explanation / Answer

Part A

What is the pH at the equivalence point?

equivalence point is when the pH changes drstically, since mol of acid = mol of base

so

this point is approx 9

Part B

What is the volume of added base at the equivalence point?

The volume at pH 8 is V = 30 mL approx

Part C

At what volume of added base is the pH calculated by working an equilibrium problem based on the initial concentration and Ka of the weak acid?

At V = 0 mL, we need to use:

Ka = [H+][A-]/[HA ]

the initial concentration of acid is used in HA

Part D

At what volume of added base does pH=pKa?

When the volume is HALF the equivalence point, since

pH =pKa + log(A-/HA)

no that the half point will have:

A- = HA

making the equation log(1)

so

pH = pKa at half equivalence point V = 15mL

Part E

At what volume of added base is the pH calculated by working an equilibrium problem based on the concentration and Kb of the conjugate base?

V base for pH equilibirum based on Kb,

this is the equivalence point

since no more acid is left, all acid has turned to weak base so we must use Kb

V = 30 mL,

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