Question

A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH.

A solution is made by mixing 9.00 mmol (millimoles) of HA and 3.00 mmol of the strong base. This resulted in a pH of 4.95.

More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 77.0 mL ?

Express the pH numerically to two decimal places.

Please help!! Please show steps if possible. THANK YOU!!

Answer 1

To do this, we need the concentration of both the acid and base, as well as the volume of the acid used. without these data, I cannot give you a numerical answer. However, I can give you guidelines for you to get the correct answer:

1. Calculate the moles of the acid (Moles = M * V) and the moles of the base.

2. Set up a first reaction between the acid and base:
HA + OH^- ------> A^- + H₂O

3. Set an ICE chart:
HA + OH^- ------> A^- + H₂O
i. moles of HA and OH 0 of A
e. moles of HA and OH consumed, moles of A = moles of OH

4. In the equivalence point all the acid is consumed and the base begins to act, so the species in solution would be the conjugate base of the weak acid.

5. Set up a second reaction of the hydrolisis of the cnjugate base:
A^- + H₂O ---------> HA + OH^-

6. Calculate the Kb = Kw/Ka

7. Set up another ICE chart with these reaction.
8. Solve for x (Concentration of OH-) by writting the equilibrium expression for this reaction

9. Calculate the pOH (pOH = -log[OH^-]

10. Calculate pH: pH = 14-pOH

Hope this guide helps you. [Post again your question with your missing data if you wish a numerical answer.