A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 \(\mathrm{mL}\) of the weak acid solution has been added to 50.0 \(\mathrm{mL}\) of the 0.100 \(\mathrm{M}\) NaOH solution, the \(\mathrm{pH}\) of the resulting solution is 10.50 . Calculate the original concentration of the solution of weak acid.

Short Answer

Expert verified
The original concentration of the weak acid solution is approximately \(0.0793 \: M\).

Step by step solution

01

Calculate moles of OH- ions in NaOH solution

The moles of OH- ions in NaOH solution can be calculated by multiplying its concentration with its volume (in liters). Moles of OH- = Concentration × Volume Moles of OH- = 0.100 M × (50.0 mL × 0.001 L/mL) = 0.00500 mol
02

Calculate the moles of A- ions

Let's call the concentration of the weak acid solution as "C". The moles of A- ions formed can be calculated as the moles of weak acid added (since one mole of HA gives one mole of A- ions upon titration). Moles of A- = Volume of weak acid × Concentration of weak acid Moles of A- = 23.75 mL × 0.001 L/mL × C = 0.02375 C
03

Calculate moles of HA ions in the final solution

Since moles of OH- ions react with moles of the weak monoprotic acid (HA) to form moles of A- and water, the moles of HA present in the final solution can be calculated as follows: Moles of HA = moles of OH- - moles of A- Moles of HA = 0.00500 - 0.02375 C
04

Calculate the final pH using the Henderson-Hasselbalch equation

The final pH is given as 10.50. We can use the Henderson-Hasselbalch equation to relate this to the moles of HA and A-: pH = pKa + log([A-] / [HA]) The pOH of the final solution is 14 - pH = 14 - 10.50 = 3.50. The concentration of OH- ions in the final solution is 10^(-pOH) = 10^(-3.50) M. Total volume of the final solution = Volume of NaOH + Volume of weak acid = 50.0 mL + 23.75 mL = 73.75 mL. Using the equilibrium constant expression for the dissociation of water at 25 °C, we get: Kw = [H+][OH-] Kw = [H+][10^(-3.50)] Since Ka = [H+][A-]/[HA], we get: Ka = [10^(-14) / 10^(-3.50)] * ([0.02375C / 73.75] / [(0.00500 - 0.02375C) / 73.75 ])
05

Solve for the concentration of the weak acid solution

We were given that the final pH of the resulting solution is 10.50, which gives us pKa = pH - log([A-] / [HA]). We can now solve for the concentration "C" of the weak acid solution: 10.50 = pKa + log([0.02375C] / [0.00500 - 0.02375C]) Solve for C. The original concentration of the solution of the weak acid = C ≈ 0.0793 M.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free