Question

$50ml,0.6Maq.$ solution of $BaC{l}_2$ is mixed with $50ml,0.6MN{a}_{3}P{O}_4$ solution at ${25}^{\circ }C$, then concentration of $B{a}^{2+}$ ions in the solution at ${25}^{\circ }C$ is:
$(K_{sp}$ of barium phosphate at ${25}^{\circ }C=2.7\times {10}^{-10})$.

• A. $1\times {10}^{-3}M$
• B. $2\times {10}^{-3}M$
• C. $3\times {10}^{-3}M$
• D. $3\times {10}^{-2}M$

Answer 1

Answer: (A)

$1\times {10}^{-3}M$

$50ml,0.6MBaC{l}_2⟶n=30millimoles$

$50ml,0.6MN{a}_{3}P{O}_4⟶n=30millimoles$

$\left[B{a}^{2+}\right]=K_{sp}\left(B{a}_3\right(P{O}_4\left)\right)=2.7\times {10}^{-10}$

$3BaCl+2N{a}_{3}P{O}_4\rightleftharpoons B{a}_3(P{O}_4{)}_2+4NaCl3030\times$

$\frac{30}{3}<\frac{30}{2}⟹\frac{x}{1}=\frac{30}{3}$

$x=10millimoles$

Total volume$=100ml$

$\left[B{a}^{2+}\right]=\frac{{10}^{-4}moles}{100\times {10}^{-3}L}={10}^{-3}M$