# Finding ionic radii from solubility data and density

$$\pu{5.35 g}$$ 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?

• 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 – MaxW Sep 28 at 15:39
• Yup I know it is CsCl structure but can you give full solution – Sameer Maurya Sep 29 at 7:19