Question

At ${25}^o$C, required volume of water, to dissolve $1$g $BaS{O}_4$ $(K_{sp}=1.1\times {10}^{-10})$ will be:

[Molecular weight of $BaS{O}_4=233$]

• A. $820L$
• B. $1L$
• C. $205L$
• D. $409L$

Answer 1

The correct option is D $409L$

We know,

$K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$

$BaS{O}_4\rightleftharpoons B{a}^{2+}+S{O}_4^{2-}$

Concentration $x$ $x$

$K_{sp}=\left[x\right]\left[x\right]$

$1.1\times {10}^{-10}=x^2$

$x=1.05\times {10}^{-5}moles/liter$

Now,

In $1liter,1.05\times {10}^{-5}moles$

In $xliters,\frac{1}{233}moles$

$x=\frac{1}{1.05\times {10}^{-5}\times 233}$

$x=408.7liters$

Hence, the correct option is $D$