Respuesta :
Answer:
[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.
Explanation:
K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)
The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.
Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.
Assuming all PbCrO₄ formed in the 175 ml solution remained ionized, then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.
To test if saturation occurs, Qsp must be > than Ksp ...
=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 => This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation. That is ...
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³
C(i) --- 0.00M 0.0336M
ΔC --- +x +x
C(eq) --- x 0.0336 + x ≅ 0.0336M
Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³
∴[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.
Answer:
[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.
Explanation:
K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)
The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.
Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.
By assumption if we say, all PbCrO₄ formed in the 175 ml solution remained ionized,
Then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.
To test if saturation occurs,
Qsp must be > than Ksp ...
=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 =>
This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation.
i.e .........
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³
C(i) --- 0.00M 0.0336M
ΔC --- +x +x
C(eq) --- x
0.0336 + x ≅ 0.0336M
Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³
Therefore;
[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.
