A buffer solution contains 0.479 M C6H5NH3Cl and 0.352 M C6H5NH2 (aniline). Determine the pH change when 0.102 mol NaOH is added to 1.00 L of the buffer.

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 =

sry but any answer choices

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