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Chapter 2: Problem 8

20.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is placed in a beaker and titrated with a solution of \(1.0 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) resulting in the creation of a precipitate. What will happen to the conductivity of the solution after additional \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is added past the equivalence point? (A) The conductivity will increase as additional ions are being added to the solution. (B) The conductivity will stay constant as the precipitation reaction has gone to completion. (C) The conductivity will decrease as the solution will be diluted with the addition of additional \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\). (D) The conductivity will stay constant as equilibrium has been established.

Short Answer

Expert verified
The correct answer is (A). After the equivalence point, adding, more Ca(NO3)2, will still contribute to the solution's ionic concentration and hence to the solution's conductivity.

Step by step solution

01

Break down the reaction

For the initial solution, 20.0 mL of 1.0 M Na2CO3 solution will break down, giving: 2Na+ + CO32-. The Na+ and CO32- ions will contribute to the conductivity of the solution.
02

Investigate the titration reaction

During the titration process, Ca(NO3)2 will react with Na2CO3 to form CaCO3, a solid precipitate, and 2NO3-. On reaching the equivalence point, all of the Na2CO3 has reacted, leaving no CO32- ions. The solution now contains Na+ and NO3- ions contributing to the conductivity.
03

Analyze the effect of additional Ca(NO3)2

When more Ca(NO3)2 is added after equivalence point, the precipitation reaction will not continue as there are no CO32- ions left. However, the additional Ca(NO3)2 will continue to produce Ca2+ and NO3- ions which will contribute to the conductivity of the solution.

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