(a) Silver sulfate, Ag2SO4, is sparingly soluble in water - CIE - A-Level Chemistry - Question 4 - 2015 - Paper 1
Question 4
(a) Silver sulfate, Ag2SO4, is sparingly soluble in water. The concentration of its saturated solution is 2.5 × 10^-2 mol dm^-3 at 298K.
(i) Write an expression for... show full transcript
Worked Solution & Example Answer:(a) Silver sulfate, Ag2SO4, is sparingly soluble in water - CIE - A-Level Chemistry - Question 4 - 2015 - Paper 1
Step 1
Write an expression for the solubility product, Ksp, of Ag2SO4 and state its units.
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Answer
The solubility product expression for Ag2SO4 is given by:
Ksp=[Ag+]2[SO42−]
The units for Ksp are mol³ dm^{-9}.
Step 2
Calculate the value for Ksp(Ag2SO4) at 298K.
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Answer
Given the concentration of the saturated solution, [Ag^+] = 2.5 × 10^-2 mol dm^-3.
We can substitute this value into the Ksp expression:
Ksp=(2.5imes10−2)2imes(2.5imes10−2)
Calculating the value:
Ksp=6.25imes10−5
Step 3
Using Ag2SO4 as an example, complete the following Hess' Law energy cycle.
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Lattice energy, ΔH_latt, is the energy required to break the ionic solid into gaseous ions:
ightarrow 2Ag^+(g) + SO_4^{2-}(g)$$
2. The enthalpy change of solution, ΔH_sol, describes the dissolution of the ionic solid in water:
$$ ext{Ag}_2SO_4(s)
ightarrow 2Ag^+(aq) + SO_4^{2-}(aq)$$
3. The enthalpy change of hydration, ΔH_hyd, represents the energy released when ions are solvated:
$$2Ag^+(g) + SO_4^{2-}(g)
ightarrow 2Ag^+(aq) + SO_4^{2-}(aq)$$
Arrows should connect these enthalpy changes appropriately.
Step 4
Use the Data Booklet to calculate the value of E°cell under standard conditions, stating which electrode is the positive one.
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Given a standard reduction potential of 0.80 V for Fe2+/Fe3+ and 0.77 V for Ag+/Ag, the cell potential is:
E°cell=E°Ag/Ag+−E°Fe2+/Fe3+=(0.80−0.77)V=0.03V
Ag+/Ag is the positive electrode.
Step 5
How would the E°cell of the above cell compare to the E°cell under standard conditions? Explain your answer.
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The E°cell would be less positive because the concentration of Ag+ in the Ag electrode is less than 1.0 mol dm^-3.
Step 6
How would the E°cell of the above cell change, if at all, if a few cm³ of concentrated Na2SO4 were added to the beaker containing Fe2+(aq) + Fe3+(aq)?
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The E°cell would decrease because the addition of concentrated Na2SO4 would increase the overall ion concentration, affecting the equilibrium of the Fe2+/Fe3+ reaction.
Step 7
How would the E°cell of the above cell change, if at all, if a few cm³ of concentrated Na2SO4 were added to the beaker containing Ag2SO4(aq)?
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The E°cell would not change because the Ag+ concentration remains relatively unaffected by the addition of Na2SO4.
Step 8
Explain any changes in E°cell you have stated in (iii).
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The increase in ion concentration from Na2SO4 shifts the equilibrium in the Fe2+/Fe3+ reaction, leading to a greater potential drop. Conversely, for Ag2SO4, the change in concentration does not significantly affect the solubility or potential.
Step 9
Calculate the pH of a 0.1 mol dm^-3 solution of iron(III) sulfate, Fe2(SO4)3.
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The concentration of H+ from the equilibrium is calculated as follows:
