Question

$80.0\text{g}$ salt of weak base and strong acid $XY$, is dissolved in water and formed $2$ liter of aqueous solution. The $pH$ of the resultant solution was found to be $5$ at $298\text{K}$. If $XY$ forms $CsCl$ type crystal having $r_{X^{+}}\left(\text{radius of}{X}^{+}\right)=1.6A^{\circ }$ and $r_{Y^{+}}\left(\text{radius of}{Y}^{+}\right)=1.864A{}^{\circ }$, then select the correct statement (s).
(Given : $K_b\left(XOH\right)=4\times {10}^{-5};N_A=6\times {10}^{23}$)

• A. Molar mass of salt is $100\text{g/mol}$
• B. $\%$ Degree of dissociation of salt is $0.25$
• C. Edge length of $XY$ is $4A^{\circ }$
• D. Density of solid salt $XY$ is $2$ in $\text{g/cc}$

Answer 1

Answer: (A), (C)

Edge length of $XY$ is $4A^{\circ }$

(A) According to questions :
$XY(aq.)\to X_{\left(aq\right)}^{+}+Y^{-}\left(aq\right)$
Therefore,
$$\begin{array}{cccc}& X_{\left(aq\right)}^{+}+& H_{2}O\left(l\right)\rightleftharpoons XOH(aq.)+& H^{+}(aq.)\\ & C(1-\alpha )& C\alpha & C\alpha \end{array}$$
$\therefore \left[H^{+}\right]=C\alpha$
Here, $K_h=\frac{(C\alpha {)}^2}{C(1-\alpha )}$
Since $\alpha$ will be very small compared to $1$, thus
$K_{h}C=(C\alpha {)}^2$
$\therefore \left[H^{+}\right]=\sqrt{\frac{K_w}{K_b}\times C}$
$\text{Again}-{\log }_{10}\left[H^{+}\right]=pH$
$\Rightarrow -{\log }_{10}\left[H^{+}\right]=5$

$C=\frac{\frac{\text{mass of salt}}{\text{Molar mass of salt (m)}}}{\text{volumes (in L)}}$
$\therefore C=\frac{80}{M\times 2}$
$k_w={10}^{-14}$
$K_b=4\times {10}^{-5}\text{( given )}$
On substititing the value in formula
we get :
$\Rightarrow {10}^{-5}=\sqrt{\frac{{10}^{-14}}{4\times {10}^{-5}}\times \frac{80}{M}\times \frac{1}{2}}$
$\Rightarrow M=\frac{{10}^{-14}\times 10}{{10}^{-5}\times {10}^{-10}}=100\text{g/ mol}$
$\text{Therefore, molar mass}=100\text{g/mol.}$

(B)
$K_w=K_a\times K_b$

$\Rightarrow K_h=K_w/K_b$

$K_h=\frac{{10}^{-14}}{4\times {10}^{-5}}=0.25\times {10}^{-9}$
As we know $K_h=C{\alpha }^2$
Therefore, $\%$ of degree of dissociation $\left(\alpha \right)$ of salt
$=\sqrt{\frac{K_h}{C}}\times 100=\sqrt{\frac{0.25\times {10}^{-9}}{0.4}}\times 100=2.5\times {10}^{-3}$

(C) $XY$ from $CsCl$ type crystal
$(r^{+}+r^{-})=\frac{\sqrt{3a}}{2}$
On putting given value in formula
$(1.6+1.864)A^{\circ }=\frac{1732}{2}\times a$
$a=\frac{2\left(3.464\right)}{1.732}=\frac{6.928}{1.732}=4A^{\circ }$

(D) Density of solid salt $XY=d=\frac{Z.M}{a^3\times N_A}$
$(Z=2\text{for CsCl type crystal})$
$=\frac{2\times 100}{(4\times {10}^{-8}{)}^3\times 6\times {10}^{23}}$
$=\frac{2\times 100}{384}$
$\text{Density (d)}=5.2\text{g/cc}$