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An aqueous solution contains dissolved c6h5nh3cl and c6h5nh2. the concentration of c6h5nh2 is 0.53 m and ph is 4.17. (a) calculate the concentration of c6h5nh3+ in this buffer solution.

Respuesta :

brownskinnjhene08

Answer: pOH = 14 - 4.22 =9.78  

Kb of aniline = 4.0 x 10^-10  

pKb = 9.40  

9.78 = 9.40 + log [C6H5NH3+] / 0.37  

10^0.38= 2.40 = [C6H5NH3+]/ 0.37  

[C6H5NH3+]= 0.89 M  

moles aniline = 0.37 M x 1 L = 0.37  

moles anilinium = 0.89 x 1 L = 0.89  

moles NaOH = 4.6 g / 40 g/mol=0.12  

C6H5NH3+ + OH- >> C6H5NH2 + H2O  

moles anilinium = 0.89 - 0.12 =0.77  

moles aniline = 0.37 + 0.12 = 0.49  

pOH = 9.40 + log 0.77/ 0.49=9.60  

pH = 4.40

Explanation:

pstnonsonjoku

The concentration of the conjugate acid of pyridine is 1.4 M

What is concentration?

The term concentration refers to the amount of substance present in solution.

We know that the pKa of C6H5NH3 + is 4.6 and the we have to use the Henderson–Hasselbalch equation.

Therefore, we have;

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

4.17 = 4.6 + log ( 0.53)/[C6H5NH3 +]

4.17 - 4.6 =  log ( 0.53)/[C6H5NH3 +]

-0.43 =  log ( 0.53)/[C6H5NH3 +]

Antilog (-0.43) = ( 0.53)/[C6H5NH3 +]

0.37 =  ( 0.53)/[C6H5NH3 +]

[C6H5NH3 +] = 0.53/0.37

[C6H5NH3 +] = 1.4 M

Learn more about pH: https://brainly.com/question/4083236

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