Question

2.7 g of a salt $ACl$ (of weak base AOH) is dissolved in 250 mL of water. The pH of the resultant solution was found to be 5. Find the density if $ACl$ forms $CsCl$ type crystal structure.
Given: $K_{b\left(AOH\right)}=2\times {10}^{-5};a^3=3\times {10}^{-23}c{m}^3;N_a=6\times {10}^{23}$

Answer 1

$ACl\to A^{+}(aq.)+C{l}^{-}(aq.)$
$A^{+}(aq.)+H_{2}O\leftrightharpoons AOH(aq.)+H^{+}(aq.)$
t = $t_{eq}$
C$(1-\alpha )$ C$\alpha$ C$\alpha$

For salts of weak base and strong acid, $\alpha ={\left(\frac{K_w}{K_b\times C}\right)}^{\frac{1}{2}}$
$[H{]}^{+}=C\alpha ={\left(\frac{K_w\times C}{K_b}\right)}^{\frac{1}{2}}$
$[H{]}^{+}=C\alpha ={\left(\frac{K_w\times w}{K_b\times M\times V\left(L\right)}\right)}^{\frac{1}{2}}$

${10}^{-5}={\left(\frac{{10}^{-14}\times 2.7\times 1000}{2\times {10}^{-5}\times M\times 250}\right)}^{\frac{1}{2}}$

${10}^{-10}=\frac{{10}^{-14}\times 2.7\times 1000}{2\times {10}^{-5}\times M\times 250}$

$M=\frac{108}{2}$

$M=54gmo{l}^{-1}$

For $CsCl$ type structure,
$\rho =\frac{ZM}{a^3\times N_a}$

$\rho =\frac{1\times 54}{3\times {10}^{-23}\times 6\times {10}^{23}}$

$\rho =3gc{m}^{-3}$