Question 10 of 10 Map A Sapling Learning macmillan learning A buffered solution

Question

Question 10 of 10 Map A Sapling Learning macmillan learning A buffered solution containing dissolved aniline, C6H5NH2, nd aniline hydrochloride, C6H5NH3Cl, has a pH of 5.32. a) Determine the concentration of C6H5NH3 in the solution if the concentration of C6H5NH2 is 0.290 M. The pKb of aniline is 9.13. Number M H. NH b) Calculate the change in pH of the solution, ApH, if 0.399 g NaOH is added to the buffer for a final volume of 1.60 L. Assume that any contribution of NaOH to the volume is negligible. Number ApH Previous Give Up & View Solution Check Answer Next Exit Hint

Explanation / Answer

a) The reaction will be

for buffer solutions we use Hendersen Hasselbalch equation which is

pH = pKa + log [salt] / [acid]

Or

pOH = pKb + log [salt] / [base]

given:

pKb = 9.13

[Base] = 0.290

pH = 5.32 therefore ,pOH = 14-pH = 14 - 5.32 = 8.68

Putting values

8.68 = 9.13 + log [salt] / 0.290

-0.45 = log [salt] / 0.290

Taking antilog

0.355 = [salt] / 0.290

[Salt] = 0.103 M

b) The Moles of NaOH added = MAss / Molecular weight = 0.399/ 40 = 0.0099 moles

Moles of base present = Molarity X volume = 0.290 X 1.60 = 0.464 moles

Moles of salt present = Molarity X volume = 0.103 X 1.60 = 0.165 moles

the moles of NaOH added will react with equal moles of salt [aniline hydrochlorde] to form equal moles of more base [aniline]

C6H5NH3+Cl- + NaOH --> C6H5NH2 + NaCl + H2O

so

Moles of salt left = 0.165 - 0.0099 = 0.155 moles

Moles of base = 0.464 + 0.0099 = 0.474 moles

New pOH of buffer will be

pOH = pKb + log [salt] / [base]

pOH = 9.13 + log [0.155 / 0.474]

pOH = 8.64

pH = 14- 8.64 = 5.36

Change in pH = 5.36 - 5.32 = 0.04

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