Question

Select the single best answer.

If 18.00 mL of 0.10 M Ba(NO₃)₂ are added to 40.00 mL of 0.10 M Na₂CO₃

a precipitate of BaCO3 will form.
	a precipitate of NaNO3 will form.
	no precipitation occurs because both possible products are soluble.
	no precipitate forms because Q < Ksp.

Answer 1

final volume = 18+ 40 = 58 ml

using dilution law M1V1 = M2V2 we find final molarities for each

Ba(NO3)2   has M1 = 0.1M , V1 = 18 ml , V2 = 58 ml

hence M2 = ( 0.1 x 18/58) = 0.031

[Ba2+] = 0.031M   ( Ba(NO3)2 gives 1Ba2+ and 2NO3-)

for Na2CO3 , M2 = ( 0.1 x 40/58) = 0.069

[CO3^2-] = 0.069 M   ( since 1Na2CO3 gives 2Na+ and CO3^2-)

Q = [Ba2+] [CO3^2-] = 0.031 x 0.069 = 0.0021

Q > ksp , hence we will get preicpiatet

answer is a precipitate of BaCO3 will form