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Chemistry Question on Equilibrium

An aqueous solution contains an unknown concentration of Ba2+Ba^{2+}. When 50 mL50 \,mL of a 1 M1\, M solution of Na2SO4Na_2SO_4 is added, BaSO4BaSO_4 just begins to precipitate. The final volume is 500 mL500 \,mL. The solubility product of BaSO4BaSO_4 is 1×10−101 \times 10^{-10}. What is the original concentration of Ba2+Ba^{2+} ?

A

5×10−9M5 \times 10^{-9} M

B

2×10−9M2 \times 10^{-9} M

C

1.1×10−9M1.1 \times 10^{-9} M

D

1.0×10−10M1.0 \times 10^{-10} M

Answer

1.1×10−9M1.1 \times 10^{-9} M

Explanation

Solution

Final concentration of [SO4−−]=[50×1][500]=0.1 M\left[ SO _{4}^{--}\right]=\frac{[50 \times 1]}{[500]}=0.1\, M
KspK _{ sp } of BaSO4,BaSO _{4},
[Ba2+][SO42−]=1×10−10{\left[ Ba ^{2+}\right]\left[ SO _{4}^{2-}\right]=1 \times 10^{-10}}
[Ba2+][0.1]=10−100.1=10−9 M{\left[ Ba ^{2+}\right][0.1]=\frac{10^{-10}}{0.1}=10^{-9} \,M }

Concentration of Ba2+Ba ^{2+} in final solution =10−9 M=10^{-9}\, M

Concentration of Ba2+Ba ^{2+} in the original solution.

M1 V1=M2 V2M _{1} \,V _{1}= M _{2}\, V _{2}
M1(500−50)=10−9(500)M _{1}(500-50)=10^{-9}(500)
M1=1.11×10−9 MM _{1}=1.11 \times 10^{-9} \,M