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5.35 g of a salt Acl (of weak base AOH) ...

5.35 g of a salt Acl (of weak base AOH) is dissolved in 250 ml of solution. The pH of the resultant solution was found to be 4.827. Find the ionic radius of `A^(+)` and `Cl^(-)` if ACl forms CsCl type crystal having 2.2 g/cm3. Given `K_(b(AOH))=1.8xx10^(-5), (r^(+))/(r^(-))=0.732` for this cell unit.

Text Solution

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`"Acl"(s) to A^(+)(aq) + Cl^(-)(aq)`
`A^(+)(aq) + H_(2)O at equilibrium `C(1-alpha)`
antilog `(-4.827) = sqrt((1 xx 10^(-4))/(1.8 xx 10^(-5)) xx (5.35 xx 1000)/(M xx 250))`
M= 53.5
For CsC l type structure.
`rho = (zM)/(a^(3) xx 6.0923 xx 10^(23))/(2.2 xx 6.023 xx 10^(23))^(1//3) = 3.43 A`
`r_(A)^(+) + r_(cl)^(-) = sqrt(3)/2 xx a = 0.866 a = 0.866 xx 3.43 A = 2.97` A
`therefore R_(Å)/R_(Cl-) = 0.732`
On solving `r_(A) = 1.255` A
`r_(Cl-) = 1.715` Å
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