Determine the $pH$ of the solution obtained by mixing enough gaseous ammonia with $50\; mL$ of a $1.758\frac{g}{L}$ ($d = 1.00$) solution $S_1$ of sodium acetate so that $53.6\;mL$ of a solution $S_2$ ($d = 0.950$) will be obtained.
$NH_3$: weak base , $CH_3COO^-$: weak base
Number of moles:
$n_{CH_3COONa}$ $=$ $n_{CH_3COO^-}$ $=$ $\frac{0.05\cdot 1.758}{82.03}$ $= $ $1.07\cdot 10^{-3}\;mol$
$m_{NH_3}$ $=$ $m_{S_2}-m_{S_1}$=$0.0536\cdot 0.950-0,050\cdot1.00$ $=$ $0.00092\;g$
$n_{NH_3}$ $=$ $\frac{0,00092}{17}$ $= $ $5.4\cdot 10^{-5}\;mol$
Initial molarities in the mixture:
$c_1$ $=$ $c_{CH_3COO^-}$ $=$ $\frac{1.07\cdot 10^{-3}}{0.0536}$ $=$ $ 0.02\;M$
$c_2$ $=$ $c_{NH_3}$ $=$ $\frac{5.4\cdot 10^-5}{0.0536}$ $= $ $0.001\;M$
$pH$ of the mixture:
$pH$ $=$ $-\frac{1}{2}log(K_{b1}c_1+K_{b2}c_3)$ =
$-\frac{1}{2}log(10^{-9.25}0.02+10^{-4.8}0.001)$
$\approx 11.6$
