$\pu{5.35 g}$ of a salt $\ce{ACl}$ is dissolved in $\pu{250 ml}$ of solution. The $\mathrm{pH}$ of the resultant solution was found to be $4.85.$ Find the ionic radius of $\ce{A+}$ and $\ce{Cl-}$ if density of $\ce{ACl}$ is $\pu{2.2 g cm-3}.$ Given

$$\frac{r_+}{r_-} = 0.73 \qquad K_\mathrm{b} = \pu{2E-5}$$

So first we can find $\log C$ by using the formula

$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b} - \log C),$$

which gives $\log C = 4.3.$ Now for the density $ρ$

$$ρ = \frac{ZM}{N_0V},$$

which will give relation in $M$ and $V,$ which are lacking. But how to proceed further?

  • $\begingroup$ The pH and Kb are red herrings. What you need to consider is the density and the ionic ratio and thus what type of packing. See the radius ratio rule $\endgroup$ – MaxW Sep 28 '19 at 15:39
  • $\begingroup$ Yup I know it is CsCl structure but can you give full solution $\endgroup$ – Sameer Maurya Sep 29 '19 at 7:19

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