Question 10 of 10 Map Sapling Learning A buffered solution containing dissolved

Question

Question 10 of 10 Map Sapling Learning A buffered solution containing dissolved aniline, CaH5NH2, and aniline hydrochloride, CaH5NH3Cl, has a pH of 5.27. a) Determine the concentration of C6H5NH3 in the solution if the concentration of CeH5NH2 is 0.335 M. The pKb of aniline is 9.13. Number CCH NH b) Calculate the change in pH of the solution ApH, if 0.354 g NaOH is added to the buffer for a final volume of 1.85 L. Assume that any contribution of NaoH to the volume is negligible. Number ApH Previous & Give Up & View Solution Check Answer Next Exit 8 Hint

Explanation / Answer

from henderson hasselbalch equation ..

pOH = pKb +log (salt/ base)

we have pOH = 14-pH = 14.5.27 = 8.73

and hence substituing values

8.73 = 9.13 + log(salt/0.335)

log(salt/0.335) = -0.4

or salt = 10-0.4 X 0.335

= 0.133 M

2) a buffger has a job to resist the change in pH hence this is what it;s gonna do when we add (0.354/40) mols of NaOH to this...

or molarity of NaOH = (0.354/40) / 1.85

= 4.78 X 10-3 M

hence change is conc of acid = change in conc of base

ie ..amount of salt used from buffer to neutralise the NaOH = amout of base formed formed from the this process

so substituiong values in the equation

pOH = pKb + log(salt/base)

pOH = 9.13 + log ([0.133 - 4.78 X 10-3 ]/[0.335 + 4.78 X 10-3 ]/)

or pOH = 9.13 + log( 0.1282/ 0.339)

=8.706

or pH = 14-8.706

= 5.29

hence delta pH = 5.29-5.27

= 0.02

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