(2pts) Post-lab Questions (1pts) 1. Do you expect the solubility of Borax to increase or decrease as temperature increases? Select the option that best explains why. Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger. Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get smaller. Solubility will decrease, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get smaller. Solubility will decrease, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger. Choose... (1pts) 2. Why was it necessary to make sure that some solid was present in the main solution before taking the samples to measure Ksp? Select the option that best explains why. To make sure no more sodium borate would dissolve in solution. To ensure the dissolution process was at equilibrium. To make sure the solution was saturated with sodium and borate ions. All of the above

Answer:

1.  Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger.

2. To ensure the dissolution process was at equilibrium.

Explanation:

Given that;

ΔG°= -RTlnK

and

ΔG° = ΔH° - TΔS°

So;

-RTlnK = ΔH° - TΔS°

lnK = ΔH°/-RT - TΔS°/-RT

lnK = -(ΔH°/RT) + ΔS°/R

K = e^-(ΔH°/RT) + ΔS°/R

Hence, Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger.

2.

Since solubility is an equilibrium process, it means that some undissolved solute must be present in order to determine the solubility product correctly.

Solubility is the ability of the solute to dissolve in the solution. As temperature increases, the solubility increases, and the presence of solid ensure dissolution.

The Gibbs free energy is the change in the energy of a system from the energy present in the reactant towards the product formation.  

It is given as,

[tex]\rm \Delta G^{\circ}= \rm -RTlnK \\\\\rm \Delta G ^{\circ} = \Delta H ^{\circ} - T\Delta S ^{\circ}[/tex]

Equating both equations we get:

[tex]\begin{aligned} \rm -RTlnK &= \rm \Delta H ^{\circ} - T\Delta S ^{\circ}\\\\\rm lnK &=\rm -(\dfrac{\Delta H ^{\circ}}{RT}) + \dfrac{\Delta S ^{\circ}}{R}\\\\\rm K &=\rm e^{-(\frac{\Delta H ^{\circ}}{RT}) + \frac{\Delta S ^{\circ}}{R}}\end{aligned}[/tex]

Thus, the solubility of the substance increases with an increase in temperature.

The solubility product is estimated correctly by the equilibrium process that ensures the presence of the undissolved solute particles.

Therefore, 1. option A. solubility increases and 2. option B. to ensure the equilibrium process are the correct options.

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