Question

Consider the titration of a 33.0 mL sample of 0.175 M HBr with 0.210 M KOH. Determine each of the following:

Part A the initial pH

Express your answer using three decimal places. pH =

Part B the volume of added base required to reach the equivalence point

Express your answer in milliliters. V =

Part C the pH at 11.0 mL of added base

Express your answer using three decimal places. pH =

Part D the pH at the equivalence point

Express your answer as a whole number. pH =

Part E the pH after adding 5.0 mL of base beyond the equivalence point

Express your answer using two decimal places pH =

Answer 1

The acid-base titration reaction that we must study in this case is:

Where HBr and KOH are strong acid and strong base, respectively.

PART A. Initial pH

The initial pH value is given only by the dissociation of HBr because we have no added titrant yet. This is:

Notice that for every mol of HBr dissolved, we have a mol of H₃O⁺ in the aqueous solution, then we can assume that:

The initial pH is given by

PART B. VOLUME OF ADDED BASE TO REACH THE EQUIVALENCE POINT

In the equivalence point, the number of moles of base react completely with the number of moles of acid. Then we can apply the equation:

Solving for V_KOH:

PART C. pH AT 11.O mL ADDED OF BASE

Before of the equivalence point, the concentration of [H₃O⁺] will decrease with the addition of base. We can use the equation:

Substituting data:

And pH is given by:

PART D: pH AT THE EQUIVALENCE POINT

In the equivalence point, the number of moles of base react completely with the number of moles of acid and KBr salt is completely formed. The value of pH is given by the dissociation of the salt. As KBr is a neutral salt, the pH in the equivalence point is given by the pure water, and it is:

PART E. pH AFTER ADDING 5.0 mL OF BASE BEYOND THE EQUIVALENCE POINT

Beyond of the equivalence point, the pH value is determined by the excess of base. We can calculate the [OH^-] concentration by using:

Substituting the known data:

The pOH is given by:

And using the relationship:

Solving for pH, we have: