Ionic solid $N a^{+} A^{-}$ crystallizes in rock salt type structure. $2.592 g$ of the ionic solid salt NaA is dissolved in water to make a 2 litre solution. The pH of this solution is 8. If the distance between the cation and the anion is $300 pm$ , calculate the density of the ionic solid $({in g/cm}^{3})$
(Given: $p K_{w} = 13, p K_{a} [H A] = 5, N_{A} = 6 \times 10^{23})$

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Solution

Step 1: Calculating the concentration of salt
When a salt undergoes hydrolysis. We can calculate the concentration of salt using the formula used for pH calculation.
$p H = \frac{1}{2} (p K_{w} + p K_{a} + l o g (c))$
Given:
$p H = 8$
$p K_{w} = 13$
$p K_{a} = 5$
$⟹ 16 = 13 + 5 + l o g (c)$
$⟹ - 2 = l o g (c)$
$⟹ c = 10^{- 2} M$
Step 2: Calculating the molar mass of salt
mass of salt used $= 2.592 g$
Volume of the solution $= 2 L$
$Molarity = \frac{Given mass}{Molar mass \times Volume of solution(L)}$
Hence, $10^{- 2} = \frac{2.592}{2 \times Molar mass}$
$Molar mass = 129.6 g / m o l$
Step 3: Calculating the density of solution
For rock salt type structure, $Z = 4$
$a = 2 \times 300 = 600 p m$
Now, density = $\frac{Z \times m a s s}{a^{3} \times N_{A}}$
$⟹ d = \frac{4 \times 129.6}{(6 \times 10^{- 8})^{3} \times 6.022 \times 10^{23}} = 3.98 g / c c$
Hence, density $= 3.98 g / c c$

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