Question

Question 1 1 pts The Ksp value for BaSO4 Ksp = 1.1 x 10 10. If 50.0 mL of 2.0 x 10-4 M Ba(NO3)2 is mixed with 350.0 mL of 1.0 x 10-5 M Na2SO4, will a precipitate be observed? A precipitate forms because the ion pair product, Q> Ksp A precipitate forms because the ion pair product, Q< Ksp. No precipitate forms because the ion pair product, Q = Ksp No precipitate forms because the ion pair product, Q<Ksp No precipitate forms because the ion pair product, Q> Ksp

Answer 1

Lets find the concentration after mixing for Ba(NO3)2

Concentration after mixing = mol of component / (total volume)

M(Ba(NO3)2) after mixing = M(Ba(NO3)2)*V(Ba(NO3)2)/(total volume)

M(Ba(NO3)2) after mixing = 2.0E-4 M*50.0 mL/(50.0+350.0)mL

M(Ba(NO3)2) after mixing = 2.5*10^-5 M

Lets find the concentration after mixing for Na2SO4

Concentration after mixing = mol of component / (total volume)

M(Na2SO4) after mixing = M(Na2SO4)*V(Na2SO4)/(total volume)

M(Na2SO4) after mixing = 1.0E-5 M*350.0 mL/(350.0+50.0)mL

M(Na2SO4) after mixing = 8.75*10^-6 M

So, we have now

[Ba2+] = 2.5*10^-5 M

[SO42-] = 8.75*10^-6 M

At equilibrium:

BaSO4 <----> Ba2+ + SO42-

Qsp = [Ba2+][SO42-]

Qsp = (2.5*10^-5)*(8.75*10^-6)

Qsp = 2.188*10^-10

we have,

Ksp = 1.1*10^-10

Since Qsp is greater than ksp, precipitate will form

Answer: option 1