Answer:
ΔpH = 0.05
Explanation:
The pKa of aniline is 4.6.
Using H-H equation for aniline, we can find the pH of the buffer before the addition of NaOH:
pH = pKa + log [C6H5NH2] / [C6H5NH3Cl]
pH = 4.6 + log [0.352] / [0.479]
pH = 4.47
Now, NaOH will react with C6H5NH3Cl as follows:
NaOH + C6H5NH3Cl → C6H5NH2 + NaCl + H₂O
That means, the moles of C6H5NH3Cl after the reaction are its initial moles - moles added of NaOH and moles of C6H5NH2 are initial moles + moles NaOH
In 1.00 of buffer, the moles of C6H5NH3Cl are 0.479 moles and moles of C6H5NH2 are 0.352 moles. The moles after the reaction are:
C6H5NH3Cl = 0.479 mol - 0.102mol = 0.377mol
C6H5NH2 = 0.352mol + 0.102 mol = 0.454mol
The pH will be:
pH = 4.6 + log [0.377mol] / [0.454mol]
pH = 4.52
And the change in pH will be:
4.52 - 4.47 =
0.05
