Determination of pH

Question

You are testing a newly discovered pH indicator which has a K_a of 2.5  10^-7. You carefully prepare a solution of 0.0041 M indicator in solutions of 0.5 M HCl, 0.5 M NaOH, and in unknown solution A. Then you put an undetermined amount of indicator into unknown solution B. You measure the visible absorption spectra of each of the four solutions. The data is illustrated in the graph and given in the table.

What is the pH of solution A, which contains 0.0041 M indicator?

What is the pH of solution B, which contains an undetermined amount of indicator?

λ (nm), ABSacid , ABSbase ,ABSsoln A, ABSsoln B

400 0.359 0.187 0.290 0.278

410 0.476 0.190 0.362 0.359

420 0.603 0.197 0.440 0.447

430 0.723 0.206 0.516 0.531

440 0.819 0.219 0.579 0.599

450 0.874 0.237 0.619 0.640

460 0.878 0.261 0.631 0.647

470 0.832 0.290 0.615 0.620

480 0.746 0.325 0.577 0.566

490 0.634 0.366 0.527 0.496

500 0.517 0.413 0.475 0.423

510 0.409 0.464 0.431 0.357

520 0.321 0.519 0.400 0.305

530 0.256 0.576 0.384 0.269

540 0.213 0.631 0.380 0.249

550 0.188 0.683 0.386 0.240

560 0.175 0.729 0.397 0.239

570 0.171 0.767 0.410 0.242

580 0.172 0.794 0.421 0.247

590 0.176 0.809 0.429 0.252

600 0.181 0.810 0.432 0.255

610 0.186 0.797 0.430 0.257

620 0.192 0.770 0.423 0.257

630 0.198 0.731 0.412 0.255

640 0.205 0.681 0.395 0.251

650 0.211 0.623 0.376 0.246

Answer 1

I think the answer this question is a lot simpler that it appears at first glance. The key is to look at the absorbance maximum for each condition. For the ABSacid sample the maximum absorbance is at 460nm (value = 0.878) for ABSbase its at 600nm (value = 0.810) for ABSsoln A and ABSsoln B the absorbance max for both solutions is at 460nm (values = 0.631 or 0.647). So that means both of the unknown solutions are at the same pH as the ABSacid sample since they share the same absorbance maxima (460nM).

So then you just need to calculate the pH of a 0.5M HCl solution

pH = -log10(H+), since HCl is a strong acid the H+ concentration is the same as the HCl concentration.

pH = -log10(0.5) = 0.301

pH = 0.301