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wolverine [178]
4 years ago
11

A student does not filter his/her saturated solution before titrating. will the calculated ksp probably be too high, too low, or

unaffected? why?
Chemistry
1 answer:
makkiz [27]4 years ago
7 0
<span>If a student does not filter his/her saturated solution before titrating then ksp value will be probably higher. To understand this, consider following titration of Ca(OH)2 with HCl.

Following reaction is initiated in above titration

Ca(OH)2 + HCl </span>→ CaCl2 + H2O

Herein, Ksp stands for solubility product. It provides the imformation of amount of solute present in solution. Now, when a sparingly solution base, like calcium hydroxide, is titrated with an acid, like HCl. The reaction results in generation of salt (in present case CaCl2) and water. The solubility of salt is higher as compared to sparingly soluble base. So during the  course of reaction, Ca^2+ ions present in system will combine with Cl^- ions to form CaCl2. This will result in decreasing in conc. of Ca^2+ ions in solution. To compensate for this lose, more Ca^2+ ions from Ca(OH)2 will dissolve in solution. Hence, Ksp value will increase. 
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The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T th
Contact [7]

Answer:

-99.8 kJ

Explanation:

We are given the methodology to answer this question, which is basically  Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.

The standard heat of reaction is

ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their  standard states.

  Compound                 ΔfHº (kJmol⁻¹)

        SO₂                             -296.8

         O₂                                    0

         SO₃                            -395.8

The balanced chemical equation is

SO₂(g) + ½O₂(g) → SO₃(g)

Thus

Δr, 298K Hº( kJmol⁻¹ ) =  1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹

Now the heat capacity of reaction  will be be given in a similar fashion:

Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants

where ν is as above the stoichiometric coefficient in the balanced chemical equation.

Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90

                         = -3.90 JK⁻¹mol⁻¹

Finally Δr,500 K Hº = Δr, 298K Hº +  CprxnΔT

Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8

                     = -99,787.8 J x 1 kJ/1000 J  = -99.8 kJ

Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.

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