Question

When 15.0 mL of a 8.00×10^-5Msodium hydroxide solution is combined with 12.0 mL of a9.32×10^-4M silver acetatesolution does a precipitate form? (yes or no)

For these conditions the Reaction Quotient, Q, is equal to.

Answer 1

NaOH(aq) + Ag(CH3COO)(aq) =>AgOH(s)+ Na(CH3COO)(aq)

Here precipitate is AgOH . It is present in solid form.

Total volume = 15+12 = 27 ml or 27*10^-3 litre

Qsp = (Ag+)*(OH-)

Moles of Ag+ = moles of Ag(CH3COO)

= 12*10^-3(in litre) * 9.32*10^-4

= 1.1184*10^-5 moles

Molarity of Ag+ = moles / total volume

1.1184*10^-5 / 27*10^-3

= 4.14*10^-4

Moles of OH- = moles of NaOH

= 15*10^-3*8*10^-5

= 1.2*10-6

Molarity = 1.2*10^-6 / 27*10^-3

= 4.44*10^-5

Now ,

Qsp = (4.44*10^-5)*(4.14*10^-4)

= 1.84*10^-8

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