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 =

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the political sponsor of weights and measures reform in the french revolutionary national assembly was the bishop of autun, better known as talleyrand. under his auspices, the french academy appointed several committees to carry out the work of developing a usable system of weights and measures for france. one of the committees recommended a decimalized measurement system based upon a length equal to one ten-millionth of the length of a quadrant of the earth’s meridian (i.e., one ten-millionth of the distance between the equator and the north pole).