Question

$\left[{Ba}^{2+}\right]$ when $Ca{SO}_4$ starts just precipitation will be:

• A. $\left[{Ba}^{2+}\right]=4.58\times {10}^{-7}M$
• B. $\left[{Ba}^{2+}\right]=5.58\times {10}^{-7}M$
• C. $\left[{Ba}^{2+}\right]=4.85\times {10}^{-7}M$
• D. None of these

Answer 1

The correct option is A $\left[{Ba}^{2+}\right]=4.58\times {10}^{-7}M$

$Ksp$ for $CaS{O}_4$ :

$K_{sp\left(CaS{O}_4\right)}=\left[C{a}^{2+}\right]\left[S{O}_4^{2-}\right]$

$2.4\times {10}^{-5}=0.1\times \left[S{O}_4^{2-}\right]$

$\left[S{O}_4^{2-}\right]=2.4\times {10}^{-4}M$

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

$1.1\times {10}^{-10}=\left[B{a}^{2+}\right]\times 2.4\times {10}^{-4}$

$\left[B{a}^{2+}\right]=4.58\times {10}^{-7}M$

Hence,option A is correct.