Question

$21.4gXCl$ salt is dissolved in water and formed $2$ litres of solution. The $pH$ of the resultant solution was found to be $5.$ Calculate the density (in $g/cc$ ) of solid salt $XCl$, if $XCl$ forms $NaCl$ type crystal having $R_{X^{+}}$(radius of $X^{+}$) $=0.8\mathring{A}$ and $R_{C{l}^{-}}$(radius of $C{l}^{-}$) $=1.7\mathring{A}$.
Given: $K_b(XOH)=2\times 10^{-5};N_A=6\times 10^{23}$ Use $10^{-0.7}=$ $0.20,\log 2=0.3$

Answer 1

Answer: 2.85

$K_{b\times OH}=2\times 10^{-5}$
$XCl\Rightarrow$ Salt of weak base and strong acid.
$pH=-\frac{1}{2}\left[\log K_w+\log C-\log K_b\right]$
$6=-\frac{1}{2}\left[\log 10^{-14}+\log C-\log \left(2\times 10^{-5}\right)\right]$
$=-\frac{1}{2}\left[-14+\log C-\log 2-\log 10^{-5}\right]$
$=-\frac{1}{2}[-14+\log C-0.3+5]$
$5=-\frac{1}{2}[-14.3+5+\log C]$
$10=14.3-5-\log C$
$\log C=14.3-5-10$
$\log C=-0.7$
$C=10^{-0.7}=0.2$
$C=0.2$
$\frac{21.4/M}{2}=0.2$
$M=\frac{21.4}{2\times 0.2}=53.5$
$a=2r^{+}+2r^{-}$
$=2\left(r^{+}+r^{-}\right)=2\times (0.8+1.7)$
$=2\times 2.5$
$a=5\mathring{A}$
$\rho =\frac{n\times \text{ mol }\text{ wt }}{N_A\times a^3}$
$\rho =\frac{4\times 53.5}{6\times 10^{23}\times {\left(5\times 10^{-8}\right)}^3}$
$=\frac{4\times 53.5}{6\times 10^{23}\times 125\times 10^{-24}}$
$=2.853$
$=2.85g/c{m}^3$