Question number 19, please show all work! Thank you!!! can be used as a pH indic

Question

Question number 19, please show all work!

Thank you!!!

can be used as a pH indicator. Suppose ti 0.001 M solution of a dye with a pKa 7.2. Froma the concentration of the protonated form is found to 0.0002 M. Assume that the remainder of the dye is in the deprotonated form. What is the pH of the solution? 6 18. What's the ratio? An acid with a p Ka of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the proton- ated to the deprotonated form of the acid? 19. Buffer capacity. Two solutions of sodium acetate are prepared, one having a concentration of 0.1 M and the other having a concentration of 0.01 M. Calculate the pH values when the following concentrations of HCl have been added to each of these solutions: 0.0025 M, 0.005 M, 0.01 M, and 0.05 M. 6

Explanation / Answer

For an acetate buffer:

apply henderson hasselbach equations

pH = pKa + log(A-/HA)

so...

a) for 0.1 M buffer:

i) 0.0025 M of acid:

[A-] = 0.1/2 = 0.05

[HA] = 0.05;

after 0.0025 M of acid:;

[A-] = 0.05-0.0025 = 0.0475

[HA] = 0.05+0.0025= 0.0525

Reclaculate in equation

pH = 4.75 + log(0.0475/0.0525) = 4.7065

ii) 0.005 M of acid:

[A-] = 0.1/2 = 0.05

[HA] = 0.05;

after 0.0050 M of acid:;

[A-] = 0.05-0.0005 = 0.0495

[HA] = 0.05+0.0050= 0.055

Reclaculate in equation

pH = 4.75 + log(0.0495/0.055) = 4.704

iii) 0.010 M of acid:

[A-] = 0.1/2 = 0.05

[HA] = 0.05;

after 0.0050 M of acid:;

[A-] = 0.05-0.010 = 0.040

[HA] = 0.05+0.010 = 0.060

Reclaculate in equation

pH = 4.75 + log(0.04/0.06) = 4.573

iv) 0.050 M of acid:

[A-] = 0.1/2 = 0.05

[HA] = 0.05;

after 0.0050 M of acid:;

[A-] = 0.05-0.05= 0

[HA] = 0.05+0.05 = 0.10

this is no longer a buffer but only acid so pH will be about 2-3

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