A buffer solution is prepared in which the concentration of $NH_3$ is 0.30 M and the concentration of $NH_4^+$

Question

A buffer solution is prepared in which the concentration of $NH_3$ is 0.30 M and the concentration of $NH_4^+$ is 0.20 M. If the equilibrium constant, $K_b$ for $NH_3$ equals $1.8 	imes 10^{-5}$, then what is the pH of this solution? ($\log 1.8 = 0.25$; $\log 0.67 = -0.176$)

Accepted Answer

According to the Henderson-Hasselbalch equation for a basic buffer :
$pOH = pK_b + \log \frac{[	ext{Conjugate acid}]}{[	ext{Base}]}$
Given:
$K_b = 1.8 	imes 10^{-5}$
$pK_b = -\log(1.8 	imes 10^{-5}) = 5 - \log 1.8 = 5 - 0.25 = 4.75$
$[	ext{Conjugate acid}] = [NH_4^+] = 0.20 	ext{ M}$
$[	ext{Base}] = [NH_3] = 0.30 	ext{ M}$
Substitute the values into the equation:
$pOH = 4.75 + \log \left(\frac{0.20}{0.30}ight) = 4.75 + \log(0.67)$
$pOH = 4.75 - 0.176 = 4.574$
We know that at 298 K, $pH + pOH = 14$ .
Therefore, $pH = 14 - pOH = 14 - 4.574 = 9.426 \approx 9.43$

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