Question

A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.58...

A buffered solution containing dissolved aniline, C6H5NH2,

and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.58

A. Determine the concentration of C6H5NH+3 in the solution if the concentration of C6H5NH2 is 0.275 M. The p?b of aniline is 9.13.

A buffered solution containing dissolved aniline, C6H5NH2,

and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.58

.

A. Determine the concentration of C6H5NH+3

in the solution if the concentration of C6H5NH2 is 0.275 M. The p?b of aniline is 9.13.

B. Calculate the change in pH of the solution, ΔpH,

if 0.398 g NaOH is added to the buffer for a final volume of 1.65 L. Assume that any contribution of NaOH to the volume is negligible.

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Answer #1

(A.) concentration of C6H5NH3+ = 0.0536 M

(B.) change in pH = 0.061

Explanation

(A.) pKb of aniline = 9.13

pKa of anilinium ion = pKw - pKa

pKa of anilinium ion = 14 - 9.13

pKa of anilinium ion = 4.87

According to Henderson - Hasselbalch equation,

pH = pKa + log([conjugate base] / [weak acid])

pH = pKa + log([C6H5NH2] / [C6H5NH3+])

5.58 = 4.87 + log(0.275 M / [C6H5NH3+])

log(0.275 M / [C6H5NH3+]) = 5.58 - 4.87

log(0.275 M / [C6H5NH3+]) = 0.71

0.275 M / [C6H5NH3+] = 100.71

0.275 M / [C6H5NH3+] = 5.13

[C6H5NH3+] = 0.275 M / 5.13

[C6H5NH3+] = 0.0536 M

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