A weak acid is titrated with 0.1236 M NaOH. From the titration curve you determine that the equivalence point occurs at 12.42 mL of added NaOH. What volume of added NaOH corresponds to the half-equivalence point?

(Incomplete question)

Assuming the molarity of the weak acid is 17.4 M, the answer would be 52.4mL.

Explanation:

Equivalence point is defined as the point where moles of titrant = moles of titrand (analyte).

At equivalence point,

# moles of NaOH = # moles of weak acid

# moles of NaOH = 0.1236 × 12.43 mL

= 0.1236 mol/L × 0.01242 L

= 0.00153511 moles of NaOH

= 0.00153511 moles of weak acid.

Since the concentration of acid is not stated in your question, we will assume the concentration of the acid to be 17.4 M.

concentration = no. of moles ÷ volume

⇒ vol. = no. of moles ÷ conc.

= 0.00153511 mol ÷ 17.4 mol/L

= 0.0267 L ≈ 26.7 mL

This means that the total volume of the solution at the half equivalence point will be:

26.7 mL + 26.7 ml

= 52.4 mL.  

N.B: Confirm missing variable from question: it could be concentration or volume of acid,but it is impossible to have two unknowns. Also, incase its pH of acid that's given, you can solve problem using Henderson-Hasslebauch equation.

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