Using the simplified formula for the $pH$ of a weak base, calculate the volume of a 19 % ($\rho_S=$ 0.9261 $\frac{g}{mL}$) ammonia solution
that must be diluted in a volumetric flask $j$ of 1000 $mL$ in order to obtain a solution with $pH$ = 11.08.
$NH_3$: weak base
Final solution:
$pOH$ $=$ $\frac{1}{2}pK_b$ $-$ $\frac{1}{2}log\;c$
$pH$ $=$ $7$ $+$ $\frac{1}{2}pK_a$ $+$ $\frac{1}{2}log\;c$
$c$ $=$ $10^{2pH$ $-$ $14-pK_a}$ =
0.09 $\frac{mol}{L}$
$n_{NH_3}$ =
$c_{NH_3}\cdot V_{j}$ =
0.09$\cdot$1 =
0.09 $\;mol$
The concentrated initial solution has the same number of moles:
Initial solution $S$:
$m_{NH_3}$ =
$n_{NH_3}\cdot M_{NH_3}$ =
0.09$\cdot$17.03 =
1.533$ \;g$
$m_S$ =
$\frac{m_{NH_3}\cdot 100}{\%_S}$ =
$\frac{1.533\cdot 100}{19}$ =
8.068$ \;g$
$V_S$ =
$\frac{m_S}{\rho_S}$ =
$\frac{1.533}{0.9261}$ =
1.655$ \; mL$
