Using the simplified formula for the $pH$ of a weak base, calculate the volume of a 22 % ($\rho_S=$ 0.9164 $\frac{g}{mL}$) ammonia solution
that must be diluted in a volumetric flask $j$ of 250 $mL$ in order to obtain a solution with $pH$ = 11.1.
$NH_3$: weak base
Final solution:
$pOH$ $=$ $\frac{1}{2}pK_b$ $-$ $\frac{1}{2}log\;c$
$pH$ $=$ $7$ $+$ $\frac{1}{2}pK_a$ $+$ $\frac{1}{2}log\;c$
$c$ $=$ $10^{2pH$ $-$ $14-pK_a}$ =
0.1 $\frac{mol}{L}$
$n_{NH_3}$ =
$c_{NH_3}\cdot V_{j}$ =
0.1$\cdot$0.25 =
0.025 $\;mol$
The concentrated initial solution has the same number of moles:
Initial solution $S$:
$m_{NH_3}$ =
$n_{NH_3}\cdot M_{NH_3}$ =
0.025$\cdot$17.03 =
0.426$ \;g$
$m_S$ =
$\frac{m_{NH_3}\cdot 100}{\%_S}$ =
$\frac{0.426\cdot 100}{22}$ =
1.936$ \;g$
$V_S$ =
$\frac{m_S}{\rho_S}$ =
$\frac{0.426}{0.9164}$ =
0.465$ \; mL$
