Question

A buffer solution is prepared in which the concentration of NH₃ is 0.30 M and the concentration of \[\ce{NH^+_4}\] is 0.20 M. If the equilibrium constant, K_b for NH₃ equals 1.8 × 10⁻⁵, what is the pH of this solution? (log 2.7 = 0.43)

Options

• 9.43

• 11.72

• 8.73

• 9.08

Answer 1

9.43

Explanation:

Given: NH₃ = 0.30 M

\[\ce{NH^+_4}\] = 0.20 M

K_b​ = 1.8 × 10⁻⁵

log 2.7 = 0.43

pK_b​ = −log(1.8 × 10⁻⁵)

= 5 − log(1.8)

≈ 5 − 0.26

= 4.74

Using the Henderson-Hasselbalch equation for a basic buffer:

\[\ce{pOH = pK_b + log \frac{[NH^4+]}{[NH3]}}\]

= \[\ce{4.74 + log \frac{0.20}{0.30}}\]

= \[\ce{4.47 + log \frac{2}{3}}\]

Now, \[\ce{log \frac{2}{3} = log \frac{1}{1.5}}\]

= −log(1.5)

≈ −0.18

pOH = 4.74 − 0.18

= 4.56

pH = 14 − pOH

pH = 14 − 4.56

= 9.44

≈ 9.43