ANALYTICAL CHEMISTRY NEED HELP WITH PART A PLEASE!!!!! What is the pH of solutio

Question

ANALYTICAL CHEMISTRY NEED HELP WITH PART A PLEASE!!!!!

What is the pH of solution A, which contains 0.0024 M indicator?

What is the pH of solution B, which contains an undetermined amount of indicator?
7.1
Copy the data to Excel and use "Chart" to plot a graph as illustrated above. Print the Excel plot and turn it in to your lab instructor, along with equations showing how you calculated the pH.

? (nm) ABSacid ABSbase ABSsoln A ABSsoln B 400 0.532 0.208 0.457 0.254 410 0.661 0.218 0.558 0.289 420 0.774 0.232 0.648 0.323 430 0.853 0.251 0.713 0.353 440 0.884 0.275 0.743 0.376 450 0.862 0.304 0.732 0.390 460 0.791 0.340 0.687 0.398 470 0.687 0.381 0.616 0.402 480 0.568 0.427 0.535 0.407 490 0.451 0.477 0.457 0.414 500 0.351 0.531 0.393 0.428 510 0.274 0.585 0.346 0.447 520 0.221 0.639 0.318 0.470 530 0.188 0.689 0.304 0.496 540 0.170 0.733 0.301 0.521 550 0.163 0.769 0.304 0.544 560 0.162 0.795 0.309 0.560 570 0.165 0.809 0.314 0.570 580 0.169 0.810 0.318 0.572 590 0.174 0.797 0.319 0.564 600 0.180 0.772 0.317 0.549 610 0.186 0.734 0.313 0.526 620 0.192 0.687 0.307 0.495 630 0.198 0.631 0.299 0.460 640 0.205 0.569 0.289 0.421 650 0.211 0.504 0.279 0.379

Explanation / Answer

(1) Ka for the indicator = 2.5 * 10^ -7

pKa= - log (2.5 * 10^ -7) = 0.3979

now we know that,

pH= pKa + log (salt)/(acid)

= 0.3979 + lg 0.5/0.5 = 0.3969

(2)- dissociation constant for indicator = sq root (Ka/C)= sq root (2.5 * 10^ -7/ 0.0024) = 1.041 * 10^ -4

concentration of B = C * dissociation constant of indicator

0.0024 = C * 1.041 * 10^ -4

C = 2.305 M

absorbance = epcilon *C* l

at 400 , 0.254 = epcilon * 0.0024 * 1

epcilon value can be further used to calculate concentration and later pH of B

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.