Question

An ionic solid $N{a}^{+}{A}^{-}$ crystallises in a rock salt type structure. 2.592 g of the ionic solid salt $N{a}^{+}{A}^{-}$is dissolved in water to make a 2 litre solution. The pH of this solution is 8. If distance between cation and anion is 300 pm. The density of the ionic solid (in gm/$c{m}^3$) is. (Given : $p{K}_w$= 13, $p{K}_a$(HA) = 5, $N_A=6\times {10}^{23}$

Answer 1

$\mathrm{Density}=\frac{Z\times M}{N_A\times a^3}$
Since it has rock salt type structure Z= 4.
As it is a rock salt type structure we can write $\frac{a}{2}=300$
$a=600$pm
a = $6\times {10}^{-8}\mathrm{cm}$
To find density we must know about molar mass,
To calculate that we can use the pH given in the question (concept of pH of a salt of strong base and weak acid) ,
$\mathrm{pH}=\frac{{\mathrm{pK}}_W}{2}+\frac{1}{2}({\mathrm{pK}}_a+\mathrm{logc})$
$8=6.5+\frac{1}{2}(5+\log \frac{2.592}{M\times 2})$
M = 129.6 gm
$\mathrm{Density}=\frac{4\times 129.6}{6\times {10}^{23}\times (6\times {10}^{-8}{)}^3}=4$