Question

Assuming no change in volume, calculate the minimum mass of $NaCl$ necessary to dissolve $0.010$ mol $AgCl$ in $100L$ solution. $\left[K_f\right(Ag{Cl}_2^{-})=3\times {10}^5$, $K_{sp}=\left(AgCl\right)=1\times {10}^{-10}]$

• A. $9.59kg$
• B. $19.5kg$
• C. $39kg$
• D. Non eof these

Answer 1

The correct option is B $19.5kg$

$AgCl+{Cl}^{-}\to {AgCl}_2^{-}$

$k_f=\frac{\left[{\mathrm{AgCl}}_2^{-}\right]}{\left[\mathrm{AgCl}\right]\left[C{l}^{-}\right]}$

$3\times {10}^5=\frac{\left[{AgCl}_2^{-}\right]}{\left[AgCl\right]\left[{Cl}^{-}\right]}$

$\left[{\mathrm{AgCl}}_2^{-}\right]$ = $\frac{\text{no of moles of AgCl}}{\text{volume}}=\frac{0.01}{100}={10}^{-4}M$

$k_{Sp}=\left[{Ag}^{+}\right]\left[C{l}^{-}\right]={10}^{-10}$

$3\times {10}^5$= $\frac{[{AgCl}_2-]}{\frac{{10}^{-10}}{\left[C{l}^{-1}\right]}{\left[C{l}^{-}\right]}^2}$

$\left[C{l}^{-}\right]=\frac{{10}^{-4}}{3\times {10}^5\times {10}^{-10}}=3.3M$

$\text{Moles of}C{l}^{-}=3.3\times 100=330moles$

$\text{Mass of}NaCl=330\times (23+35.5)$

$=19.5kg$