When 5.35 gm Acl is dissolved in 250 ml of water the pH of solution is found to be 4.827. Find the ionic radius of A+ and Cl^(-). If "Acl" forms CsCl type crystal having 2.2 g//cm^(3). Given K_(b)(AOH) = 1.8 xx 10^(-5) r^(+)/r^(-) = 0.732 for this cell unit.

When 5.35 gm Acl is dissolved in 250 ml of water the pH of solution is

1.	When 5.35 gm Acl is dissolved in 250 ml of water the pH of solution is found to be 4.827. Find the ionic radius of A+ and Cl^(-). If "Acl" forms CsCl type crystal having 2.2 g//cm^(3). Given K_(b)(AOH) = 1.8 xx 10^(-5) r^(+)/r^(-) = 0.732 for this cell unit.
Answer» SOLUTION :`"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