Abuffered solution containing dissolved aniline, c6h5nh2, and aniline hydrochloride, c6h5nh3cl, has a ph of 5.75.

a) determine the concentration of c6h5nh3+ in the solution if the concentration of c6h5nh2 is 0.245 m. the pkb of aniline is 9.13.

[c6h5nh3+]= m

b) calculate the change in ph of the solution, \delta ph, if 0.360g naoh is added to the buffer for a final volume of 1.50l.

\deltaph=

a) 0.0323 M

b) ΔpH = 0.10

Explanation:

A buffer is a solution of a weak acid and its conjugate base, or of a weak base and its conjugate acid. This solution keeps the pH almost constant because it has equilibrium between the acid and the base. So, when acid is added, it reacts with the base and when a base is added, it reacts with the acid.

To calculate the pH in a buffered solution, we may use the Henderson-Hasselbalch equation:

Where pKa is the -logKa and Ka is the acid constant, and [A⁻] is the concentration of the base (or ionic compound of the acid), and [HA] is the concentration of the acid.

a) C₆H₅NH₂ is the base and its conjugate acid is C₆H₅NH₃⁺. pKa + pKb = pKw, where pKw is from the water equilibrium and is equal to 14.

pKa = 14 - 9.13 = 4.87

So,

pH = pKa + log [C₆H₅NH₂]/[C₆H₅NH₃⁺]

5.75 = 4.87 + log (0.245)/[C₆H₅NH₃⁺]

5.75 - 4.87 = log (0.245) - log [C₆H₅NH₃⁺]

log[C₆H₅NH₃⁺] = -0.6108 - 0.88

log [C₆H₅NH₃⁺] = -1.4908

[C₆H₅NH₃⁺] =

[C₆H₅NH₃⁺] = 0.0323 M

b) First, let's calculated the concentration of NaOH, knowing that the molar masses are

MNa = 23 g/mol

MO = 16 g/mol

MH = 1 g/mol

MNaOH = 40 g/mol

n = m/M

n = 0.360/40 = 0.009 mol

The concentration is: n/V = 0.009/1.5 = 0.006 M.

The base reacts with the acid to form more base compound. Than the concnetration of aniline and aniline hydrochloride will be know:

[C₆H₅NH₂] = 0.245 + 0.006 = 0.2510 M

[C₆H₅NH₃⁺] = 0.0323 - 0.006 = 0.0263 M

So, the new pH will be:

pH = 4.87 + log(0.2510)/(0.0263)

pH = 4.87 + 0.98

pH = 5.85

ΔpH = 5.85 - 5.75

ΔpH = 0.10