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WITCHER [35]
3 years ago
13

A student conducting the iodine clock experiment accidentally makes an s2o32- stock solution that is too concentrated. how will

this affect the rate measurement? reaction 1: 3i- (aq) + s2o82- (aq) -> i3- (aq) + 2so42- (aq) slow reaction 2: i3- (aq) + 2s2o32- (aq) -> 3i- (aq) + s4o62- (aq) fast reaction 3: i3- (aq) + starch (aq) -> 3i- ---- starch (bluish black) fast
Chemistry
1 answer:
cupoosta [38]3 years ago
8 0
Adding (S2O3)2- would affect the reaction mechanism that involves this ion. From the reaction mechanism given above, the equilibrium of step 2 would be affected. Adding the stock solution of (S2O3)2- would shift the equilibrium to the right thus making more products of the said mechanism. Also, the reaction rate of this step would occur faster than the original rate. This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed. So, that after the addition, a color change would appear immediately because I3- would be removed slowly from solution, and would therefore be able to react with starch. 

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For each pair, choose the compound with the smaller lattice energy, and explain your choice: (b) K₂O or K₂S.
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the compound with the smaller lattice energy is potassium sulphide here the size of the molecule play a important role

The quantity of energy released by the electrostatic attraction of oppositely charged ions is known as lattice energy (L.E). The ion's size and charge affect the lattice energy.

lattice energy is inversely proportional to   size of ion and directly proportional to charge of the ion. They are each charged equally that is plus two and minus two However, because the Sulphur is larger in size  and the oxygen is lesser in this case, The lattice energy of potassium oxide is larger  the lattice energy of potassium sulphide  is smaller.

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The molar solubility of magnesium carbonate is 1.8 × 10–4 mol/l. What is ksp for this compound?
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Answer:

3.2 × 10⁻⁸

Explanation:

Let's consider the solution of magnesium carbonate.

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We can relate the molar solubility (S) with the solubility product (Ksp) using an ICE chart.

         MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

I                             0                0

C                          +S              +S

E                            S                S

The Ksp is:

Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸

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