Katieg3593
Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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.

Sagot :

jufsanabriasa

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

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.