Looking for classes? Ksquare Career Institute, Bengaluru →
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 $1.8\times10^{-5}$, what is the pH of this solution?
A
8.73
B
9.08
C
9.43
D
11.72
Detailed Solution
This is a basic buffer of a weak base ($NH_3$) and its salt ($NH_4^+$).
$pK_b = -\log(1.8\times10^{-5}) = 5 - \log1.8 = 5 - 0.26 = 4.74$
Henderson equation: $pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}$
$pOH = 4.74 + \log\frac{0.20}{0.30} = 4.74 + (\log2 - \log3) = 4.74 + (0.301 - 0.477)$
$pOH = 4.74 - 0.176 = 4.56$
$pH = 14 - pOH = 14 - 4.56 = 9.44$
This matches the value 9.43 among the choices.
$pK_b = -\log(1.8\times10^{-5}) = 5 - \log1.8 = 5 - 0.26 = 4.74$
Henderson equation: $pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}$
$pOH = 4.74 + \log\frac{0.20}{0.30} = 4.74 + (\log2 - \log3) = 4.74 + (0.301 - 0.477)$
$pOH = 4.74 - 0.176 = 4.56$
$pH = 14 - pOH = 14 - 4.56 = 9.44$
This matches the value 9.43 among the choices.