A quantity of 80 gm of a salt $[X Y]$ of strong acid and weak base is dissolved in water to form 2 lit aqueous solution. At 298K the pH of the solution is found to be 5. If $[X Y]$ forms $C s C l$ type crystal and the radius of $X^{+}$ and $Y^{-}$ ions are 160 pm and 186.4 pm respectively. Which of the following is/are correct ?
${[K_{b}]}_{X O H} = 4 \times 10^{- 5}, N_{A} = 6 \times 10^{23}, \sqrt{3} = 1.732$ $[\log 4 = 0.6]$

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Density of solid $X Y$ is $2.6 {g/cm}^{3}$ .

Degree of hydrolysis of salt is $2.5 \times 10^{- 5}$

Unit cell edge length of $X Y$ crystal is 400 pm

Molar mass of the salt is 100 g/mol

answer is A, B, C, D.

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Detailed Solution

$p H = 7 - \frac{1}{2} p K b - \frac{1}{2} \log c$
a) $5 = 7 - \frac{1}{2} \times 4.4 - \frac{1}{2} \log c$
$- \log c = 0.4 c = 0.4$
$0.4 = \frac{80}{G m w} \times \frac{1}{2}$
$G m w = \frac{40}{0.4} = 100 g/mole$
b) $X^{+} + H_{2} O ⇌ X O H + H^{+}$
$C O O$
$C - C h C h C h$
$[H^{+}] = C h$
$10^{- 5} = 0.4 \times h$
$h = 2.5 \times 10^{- 5}$
c) $\sqrt{3} a = 2 [r_{x^{+}} + r_{y^{-}}]$
$a = \frac{2 (160 + 186.4)}{1.732} = 400 p m$
d) $d = \frac{1 \times 100}{6 \times 10^{23} \times {(4 \times 10^{- 10})}^{3}}$
$d = 2.6 \times 10^{6} {g/cm}^{3}$

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