Select the single best answer. If 18.00 mL of 0.10 M Ba(NO 3 ) 2 are added to 40

ID: 560034 • Letter: S

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.

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.

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

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