This note explores how solubility equilibria are quantified through K s p K_{sp} K s p common ions ) dramatically suppresses solubility. We'll derive K s p K_{sp} K s p selective precipitation schemes to separate mixed ions—a cornerstone technique in qualitative analysis and industrial purification.
Intuition Why Does Adding a Common Ion Reduce Solubility?Imagine a saturated salt solution at equilibrium: solid⇌ dissolved ions. The equilibrium "resists change" (Le Chatelier). If you dump in extra product ions from another source, the equilibrium shifts backward to consume them—more solid precipitates out, fewer original salt molecules dissolve. It's like a crowded room: if the room already has many people (ions), fewer newcomers (salt molecules) can fit in.
Definition Solubility Product Constant (K s p K_{sp} K s p M x A y ( s ) ⇌ x M n + ( a q ) + y A m − ( a q ) M_xA_y(s) \rightleftharpoons xM^{n+}(aq) + yA^{m-}(aq) M x A y ( s ) ⇌ x M n + ( a q ) + y A m − ( a q ) solubility product is:
K s p = [ M n + ] x [ A m − ] y K_{sp} = [\text{M}^{n+}]^x [\text{A}^{m-}]^y K s p = [ M n + ] x [ A m − ] y Key points:
Pure solid concentration does NOT appear (activity = 1).
K s p K_{sp} K s p Units: ( mol/L ) x + y (\text{mol/L})^{x+y} ( mol/L ) x + y WHY this form? It's the equilibrium constant K c K_c K c product of ion concentrations raised to stoichiometric powers is constant.
Intuition Solubility (s) vs.K s p K_{sp} K s p
Solubility (s) : grams per liter or mol/L of salt that dissolves. It's what you measure in the lab.K s p K_{sp} K s p product of ion concentrations at saturation. It's thermodynamic fingerprint.
Relationship: For M x A y \text{M}_x\text{A}_y M x A y s s s [ M n + ] = x s [\text{M}^{n+}] = xs [ M n + ] = x s [ A m − ] = y s [\text{A}^{m-}] = ys [ A m − ] = y s K s p = ( x s ) x ( y s ) y = x x y y s x + y K_{sp} = (xs)^x (ys)^y = x^x y^y s^{x+y} K s p = ( x s ) x ( y s ) y = x x y y s x + y s s s s = ( K s p x x y y ) 1 / ( x + y ) s = \left(\frac{K_{sp}}{x^x y^y}\right)^{1/(x+y)} s = ( x x y y K s p ) 1/ ( x + y )
Worked example Example 1: Calculating Solubility fromK s p K_{sp} K s p Problem: K s p K_{sp} K s p AgCl \text{AgCl} AgCl 1.8 × 10 − 10 1.8 \times 10^{-10} 1.8 × 1 0 − 10 Solution:
Write the equilibrium: AgCl(s) ↔ Ag + (aq) + Cl − (aq) \text{AgCl(s)} \leftrightarrow \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} AgCl(s) ↔ Ag + (aq) + Cl − (aq) Define solubility: Let s s s [ Ag + ] = s [\text{Ag}^+] = s [ Ag + ] = s [ Cl − ] = s [\text{Cl}^-] = s [ Cl − ] = s Substitute into K s p K_{sp} K s p
K s p = [ Ag + ] [ Cl − ] = s ⋅ s = s 2 K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \cdot s = s^2 K s p = [ Ag + ] [ Cl − ] = s ⋅ s = s 2 1.8 × 10 − 10 = s 2 1.8 \times 10^{-10} = s^2 1.8 × 1 0 − 10 = s 2 Solve:
s = 1.8 × 10 − 10 = 1.34 × 10 − 5 mol/L s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ mol/L} s = 1.8 × 1 0 − 10 = 1.34 × 1 0 − 5 mol/L
Why this step? We equate the product of equilibrium concentrations to K s p K_{sp} K s p s s s
Worked example Example 2: Solubility of a 1:2 SaltProblem: K s p K_{sp} K s p CaF 2 \text{CaF}_2 CaF 2 3.9 × 10 − 11 3.9 \times 10^{-11} 3.9 × 1 0 − 11 Solution:
Equilibrium: C a F 2 ( s ) ↔ C a 2 + ( a q ) + 2 F − ( a q ) CaF_2(s) \leftrightarrow Ca^{2+}(aq) + 2F^-(aq) C a F 2 ( s ) ↔ C a 2 + ( a q ) + 2 F − ( a q ) Stoichiometry: If s s s [ Ca 2 + ] = s [\text{Ca}^{2+}] = s [ Ca 2 + ] = s [ F − ] = 2 s [\text{F}^-] = 2s [ F − ] = 2 s Substitute:
K s p = [ Ca 2 + ] [ F − ] 2 = s ( 2 s ) 2 = 4 s 3 K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s(2s)^2 = 4s^3 K s p = [ Ca 2 + ] [ F − ] 2 = s ( 2 s ) 2 = 4 s 3 3.9 × 10 − 11 = 4 s 3 3.9 \times 10^{-11} = 4s^3 3.9 × 1 0 − 11 = 4 s 3 Solve:
s 3 = 3.9 × 10 − 11 4 = 9.75 × 10 − 12 s^3 = \frac{3.9 \times 10^{-11}}{4} = 9.75 \times 10^{-12} s 3 = 4 3.9 × 1 0 − 11 = 9.75 × 1 0 − 12 s = ( 9.75 × 10 − 12 ) 1 / 3 = 2.14 × 10 − 4 mol/L s = (9.75 \times 10^{-12})^{1/3} = 2.14 \times 10^{-4} \text{ mol/L} s = ( 9.75 × 1 0 − 12 ) 1/3 = 2.14 × 1 0 − 4 mol/L
Why this step? The 1:2 stoichiometry means for every Ca²⁺, two F⁻ appear. We square [ F − ] [\text{F}^-] [ F − ] K s p K_{sp} K s p
Definition Common-Ion EffectAdding a soluble salt that shares an ion with a sparingly soluble salt suppresses the solubility of the sparingly soluble salt. The shared ion is the common ion .
WHY? Le Chatelier: increasing a product concentration shifts equilibrium toward reactants (the solid). Mathematically, if one ion's concentration rises, the other must fall to keep K s p K_{sp} K s p
Worked example Example 3: Common-Ion Effect with AgClProblem: Calculate solubility of AgCl in 0.10 M NaCl solution. (K s p = 1.8 × 10 − 10 K_{sp} = 1.8 \times 10^{-10} K s p = 1.8 × 1 0 − 10
Solution:
Equilibrium: A g C l ( s ) ⇌ A g + + C l − AgCl(s) \rightleftharpoons Ag^+ + Cl^- A g C l ( s ) ⇌ A g + + C l − Initial conditions: [ Cl − ] = 0.10 [\text{Cl}^-] = 0.10 [ Cl − ] = 0.10 [ Ag + ] = 0 [\text{Ag}^+] = 0 [ Ag + ] = 0 At equilibrium: [ Ag + ] = s ′ [\text{Ag}^+] = s' [ Ag + ] = s ′ [ Cl − ] = 0.10 + s ′ [\text{Cl}^-] = 0.10 + s' [ Cl − ] = 0.10 + s ′ Substitute:
K s p = s ′ ( 0.10 + s ′ ) K_{sp} = s'(0.10 + s') K s p = s ′ ( 0.10 + s ′ ) Assume s ′ ≪ 0.10 s' \ll 0.10 s ′ ≪ 0.10 Then 0.10 + s ′ ≈ 0.10 0.10 + s' \approx 0.10 0.10 + s ′ ≈ 0.10 1.8 × 10 − 10 = s ′ ( 0.10 ) 1.8 \times 10^{-10} = s'(0.10) 1.8 × 1 0 − 10 = s ′ ( 0.10 ) s ′ = 1.8 × 10 − 10 0.10 = 1.8 × 10 − 9 mol/L s' = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \text{ mol/L} s ′ = 0.10 1.8 × 1 0 − 10 = 1.8 × 1 0 − 9 mol/L Check assumption: 1.8 × 10 − 9 ≪ 0.10 1.8 \times 10^{-9} \ll 0.10 1.8 × 1 0 − 9 ≪ 0.10 Why this step? The approximation 0.10 + s ′ ≈ 0.10 0.10 + s' \approx 0.10 0.10 + s ′ ≈ 0.10 [ Cl − ] [\text{Cl}^-] [ Cl − ]
Compare: In pure water, s = 1.34 × 10 − 5 s = 1.34 \times 10^{-5} s = 1.34 × 1 0 − 5 s ′ = 1.8 × 10 − 9 s' = 1.8 \times 10^{-9} s ′ = 1.8 × 1 0 − 9 7500 times less soluble!
Worked example Example 4: Common-Ion with 1:2 SaltProblem: Solubility of CaF 2 \text{CaF}_2 CaF 2 K s p = 3.9 × 10 − 11 K_{sp} = 3.9 \times 10^{-11} K s p = 3.9 × 1 0 − 11 Solution:
Equilibrium: CaF 2 (s) ⇌ Ca 2 + + 2 F − \text{CaF}_2\text{(s)} \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^- CaF 2 (s) ⇌ Ca 2 + + 2 F − Initial: [ F − ] = 0.010 [\text{F}^-] = 0.010 [ F − ] = 0.010 [ Ca 2 + ] = 0 [\text{Ca}^{2+}] = 0 [ Ca 2 + ] = 0 At equilibrium: [ Ca 2 + ] = s ′ [\text{Ca}^{2+}] = s' [ Ca 2 + ] = s ′ [ F − ] = 0.010 + 2 s ′ [\text{F}^-] = 0.010 + 2s' [ F − ] = 0.010 + 2 s ′ Assume 2 s ′ ≪ 0.010 2s' \ll 0.010 2 s ′ ≪ 0.010 [ F − ] ≈ 0.010 [\text{F}^-] \approx 0.010 [ F − ] ≈ 0.010 K s p = s ′ ( 0.010 ) 2 = s ′ × 10 − 4 K_{sp} = s'(0.010)^2 = s' \times 10^{-4} K s p = s ′ ( 0.010 ) 2 = s ′ × 1 0 − 4 s ′ = 3.9 × 10 − 11 10 − 4 = 3.9 × 10 − 7 mol/L s' = \frac{3.9 \times 10^{-11}}{10^{-4}} = 3.9 \times 10^{-7} \text{ mol/L} s ′ = 1 0 − 4 3.9 × 1 0 − 11 = 3.9 × 1 0 − 7 mol/L Check: 2 s ′ = 7.8 × 10 − 7 ≪ 0.010 2s' = 7.8 \times 10^{-7} \ll 0.010 2 s ′ = 7.8 × 1 0 − 7 ≪ 0.010
Why this step? We squared [ F − ] [\text{F}^-] [ F − ]
Compare: In pure water, s = 2.14 × 10 − 4 s = 2.14 \times 10^{-4} s = 2.14 × 1 0 − 4 s ′ = 3.9 × 10 − 7 s' = 3.9 \times 10^{-7} s ′ = 3.9 × 1 0 − 7 549 times less soluble.
Definition Ionic Product (Q) vs.K s p K_{sp} K s p M x A y \text{M}_x\text{A}_y M x A y ionic product is:
Q = [ M n + ] x [ A m − ] y Q = [\mathrm{M}^{n+}]^x [\mathrm{A}^{m-}]^y Q = [ M n + ] x [ A m − ] y any concentrations, not necessarily at equilibrium.
Decision rule:
If Q < K s p Q < K_{sp} Q < K s p unsaturated , no precipitate, more salt can dissolve.
If Q = K s p Q = K_{sp} Q = K s p saturated , equilibrium, solid and solution coexist.
If Q > K s p Q > K_{sp} Q > K s p supersaturated , precipitation occurs until Q = K s p Q = K_{sp} Q = K s p
K s p K_{sp} K s p K s p K_{sp} K s p K s p K_{sp} K s p
Worked example Example 5: Will a Precipitate Form?Problem: Mix 100 mL of 0.010 M AgNO 3 \text{AgNO}_3 AgNO 3 K s p = 1.8 × 10 − 10 K_{sp} = 1.8 \times 10^{-10} K s p = 1.8 × 1 0 − 10 Solution:
After mixing, total volume = 200 mL. Concentrations are diluted by factor of 2.
[ Ag + ] = 0.0010 2 = 5.0 × 10 − 4 M [\text{Ag}^+] = \frac{0.0010}{2} = 5.0 \times 10^{-4} \text{ M} [ Ag + ] = 2 0.0010 = 5.0 × 1 0 − 4 M [ Cl − ] = 0.0020 2 = 1.0 × 10 − 3 M [\text{Cl}^-] = \frac{0.0020}{2} = 1.0 \times 10^{-3} \text{ M} [ Cl − ] = 2 0.0020 = 1.0 × 1 0 − 3 M Calculate Q:
Q = [ Ag + ] [ Cl − ] = ( 5.0 × 10 − 4 ) ( 1.0 × 10 − 3 ) = 5.0 × 10 − 7 Q = [\text{Ag}^+][\text{Cl}^-] = (5.0 \times 10^{-4})(1.0 \times 10^{-3}) = 5.0 \times 10^{-7} Q = [ Ag + ] [ Cl − ] = ( 5.0 × 1 0 − 4 ) ( 1.0 × 1 0 − 3 ) = 5.0 × 1 0 − 7 Compare: Q = 5.0 × 10 − 7 > K s p = 1.8 × 10 − 10 Q = 5.0 \times 10^{-7} > K_{sp} = 1.8 \times 10^{-10} Q = 5.0 × 1 0 − 7 > K s p = 1.8 × 1 0 − 10 Conclusion: Yes, AgCl will precipitate.
Why this step? We dilute because mixing increases volume. Then we compute the actual ion product and compare to the threshold. Since Q > K s p Q > K_{sp} Q > K s p
Definition Selective PrecipitationA technique to separate a mixture of cations (or anions) by adding a precipitating agent that selectively brings down one ion while leaving others in solution. The key is exploiting large differences in K s p K_{sp} K s p
Imagine two salts with the same anion but vastly different K s p K_{sp} K s p K s p K_{sp} K s p much lower anion concentration. By carefully controlling how much precipitating agent you add, you can crash out one ion, filter it off, then add more reagent to precipitate the second. It's like fishing: use the right bait concentration to catch the "low-solubility fish" first.
Worked example Example 6: Separating Ag⁺ and Pb²⁺ with ChlorideProblem: A solution contains 0.010 M Ag⁺ and 0.010 M Pb²⁺. Can you separate them by slowly adding Cl⁻? K s p ( AgCl ) = 1.8 × 10 − 10 K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10} K s p ( AgCl ) = 1.8 × 1 0 − 10 K s p ( PbCl 2 ) = 1.7 × 10 − 5 K_{sp}(\text{PbCl}_2) = 1.7 \times 10^{-5} K s p ( PbCl 2 ) = 1.7 × 1 0 − 5
Solution:
Find [ Cl − ] [\text{Cl}^-] [ Cl − ]
For AgCl: K s p = [ Ag + ] [ Cl − ] K_{sp} = [\text{Ag}^+][\text{Cl}^-] K s p = [ Ag + ] [ Cl − ] [ Cl − ] AgCl = K s p [ Ag + ] = 1.8 × 10 − 10 0.010 = 1.8 × 10 − 8 M [\text{Cl}^-]_{\text{AgCl}} = \frac{K_{sp}}{[\text{Ag}^+]} = \frac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8} \text{ M} [ Cl − ] AgCl = [ Ag + ] K s p = 0.010 1.8 × 1 0 − 10 = 1.8 × 1 0 − 8 M
For PbCl₂: K s p = [ Pb 2 + ] [ Cl − ] 2 K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 K s p = [ Pb 2 + ] [ Cl − ] 2 [ Cl − ] PbCl 2 = K s p [ Pb 2 + ] = 1.7 × 10 − 5 0.010 = 1.7 × 10 − 3 = 0.041 M [\text{Cl}^-]_{\text{PbCl}_2} = \sqrt{\frac{K_{sp}}{[\text{Pb}^{2+}]}} = \sqrt{\frac{1.7 \times 10^{-5}}{0.010}} = \sqrt{1.7 \times 10^{-3}} = 0.041 \text{ M} [ Cl − ] PbCl 2 = [ Pb 2 + ] K s p = 0.010 1.7 × 1 0 − 5 = 1.7 × 1 0 − 3 = 0.041 M
Compare: AgCl precipitates at 1.8 × 10 − 8 1.8 \times 10^{-8} 1.8 × 1 0 − 8 Strategy: Slowly add Cl⁻. When [ Cl − ] = 1.8 × 10 − 8 [\text{Cl}^-] = 1.8 \times 10^{-8} [ Cl − ] = 1.8 × 1 0 − 8 [ Cl − ] [\text{Cl}^-] [ Cl − ] Filter off AgCl. Then add more Cl⁻ past0.041 M to precipitate PbCl₂.
Why this step? We calculate the threshold [ Cl − ] [\text{Cl}^-] [ Cl − ] [ Ag + ] [\text{Ag}^+] [ Ag + ] 10 − 5 10^{-5} 1 0 − 5
Quantitative check: When [ Cl − ] = 0.041 [\text{Cl}^-] = 0.041 [ Cl − ] = 0.041 [ Ag + ] [\text{Ag}^+] [ Ag + ] [ Ag + ] = K sp [ Cl − ] = 1.8 × 10 − 10 0.041 = 4.4 × 10 − 9 M [\text{Ag}^+] = \frac{K_{\text{sp}}}{[\text{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.041} = 4.4 \times 10^{-9} \text{ M} [ Ag + ] = [ Cl − ] K sp = 0.041 1.8 × 1 0 − 10 = 4.4 × 1 0 − 9 M
Worked example Example 7: Fractional Precipitation of SulfidesProblem: A solution contains 0.10 M Cu²⁺ and 0.10 M Mn²⁺. Adjust pH to precipitate CuS but not MnS by controlling [ S 2 − ] [\text{S}^{2-}] [ S 2 − ] K s p ( CuS ) = 6 × 10 − 37 K_{sp}(\text{CuS}) = 6 \times 10^{-37} K s p ( CuS ) = 6 × 1 0 − 37 K s p ( MnS ) = 3 × 10 − 14 K_{sp}(\text{MnS}) = 3 \times 10^{-14} K s p ( MnS ) = 3 × 1 0 − 14
Solution:
Find [ S 2 − ] [\text{S}^{2-}] [ S 2 − ]
CuS: K s p = [ Cu 2 + ] [ S 2 − ] K_{sp} = [\text{Cu}^{2+}][\text{S}^{2-}] K s p = [ Cu 2 + ] [ S 2 − ] [ S 2 − ] CuS = 6 × 10 − 37 0.10 = 6 × 10 − 36 M [\text{S}^{2-}]_{\text{CuS}} = \frac{6 \times 10^{-37}}{0.10} = 6 \times 10^{-36} \text{ M} [ S 2 − ] CuS = 0.10 6 × 1 0 − 37 = 6 × 1 0 − 36 M
MnS: [ S 2 − ] MnS = 3 × 10 − 14 0.10 = 3 × 10 − 13 M [\text{S}^{2-}]_{\text{MnS}} = \frac{3 \times 10^{-14}}{0.10} = 3 \times 10^{-13} \text{ M} [ S 2 − ] MnS = 0.10 3 × 1 0 − 14 = 3 × 1 0 − 13 M
Range: Keep 6 × 10 − 36 < [ S 2 − ] < 3 × 10 − 13 6 \times 10^{-36} < [\text{S}^{2-}] < 3 \times 10^{-13} 6 × 1 0 − 36 < [ S 2 − ] < 3 × 1 0 − 13 Mechanism: [ \ce S 2 − ] [\ce{S^{2-}}] [ \ce S 2 − ] \ce H 2 S \ce{H2S} \ce H 2 S \ce H 2 S < = > 2 H + + S 2 − \ce{H2S <=> 2H+ + S^{2-}} \ce H 2 S <=> 2 H + + S 2 − [ \ce S 2 − ] [\ce{S^{2-}}] [ \ce S 2 − ] Conclusion: CuS precipitates at extremely low [ \ce S 2 − ] [\ce{S^{2-}}] [ \ce S 2 − ] 10 23 10^{23} 1 0 23
Why this step? The colosal K s p K_{sp} K s p
Common mistake Mistake 1: Forgetting Stoichiometry inK s p K_{sp} K s p Wrong approach: For \ce C a 3 ( P O 4 ) 2 \ce{Ca3(PO4)2} \ce C a 3 ( P O 4 ) 2 K s p = [ \ce C a 2 + ] [ \ce P O 4 3 − ] K_{sp} = [\ce{Ca^{2+}}][\ce{PO4^{3-}}] K s p = [ \ce C a 2 + ] [ \ce P O 4 3 − ] Why it feels right: You might think "one Ca²⁺, one PO₄³⁻" without checking coefficients.
The fix: Always write the balanced dissolution:
\ce C a 3 ( P O 4 ) 2 ( s ) < = > 3 C a 2 + + 2 P O 4 3 − \ce{Ca3(PO4)2(s) <=> 3Ca^{2+} + 2PO4^{3-}} \ce C a 3 ( P O 4 ) 2 ( s ) <=> 3 C a 2 + + 2 P O 4 3 − K s p = [ \ce C a 2 + ] 3 [ \ce P O 4 3 − ] 2 K_{sp} = [\ce{Ca^{2+}}]^3 [\ce{PO4^{3-}}]^2 K s p = [ \ce C a 2 + ] 3 [ \ce P O 4 3 − ] 2 s s s [ \ce C a 2 + ] = 3 s [\ce{Ca^{2+}}] = 3s [ \ce C a 2 + ] = 3 s [ \ce P O 4 3 − ] = 2 s [\ce{PO4^{3-}}] = 2s [ \ce P O 4 3 − ] = 2 s K s p = ( 3 s ) 3 ( 2 s ) 2 = 108 s 5 K_{sp} = (3s)^3(2s)^2 = 108s^5 K s p = ( 3 s ) 3 ( 2 s ) 2 = 108 s 5
Steel-man: The mistake arises because in simple1:1 salts (NaCl, AgCl), the exponents are all 1, so you internalize "just multiply concentrations." The fix is to always derive from the balanced equation.
Common mistake Mistake 2: Ignoring the Common-Ion Approximation BreakdownWrong approach: Using s ′ ≈ K s p / C s' \approx K_{sp}/C s ′ ≈ K s p / C s ′ s' s ′ C C C
Why it feels right: The approximation simplifies the math, and you've been told "if C C C s ′ s' s ′
The fix: After solving, check s ′ ≪ C s' \ll C s ′ ≪ C s ′ / C > 0.05 s'/C > 0.05 s ′ / C > 0.05 K s p = s ′ ( C + s ′ ) ⟹ s ′ 2 + C s ′ − K s p = 0 K_{sp} = s'(C + s') \implies s'^2 + Cs' - K_{sp} = 0 K s p = s ′ ( C + s ′ ) ⟹ s ′2 + C s ′ − K s p = 0 Steel-man: The approximation works 95% of the time in textbook problems, so you forget to verify. The fix is discipline: always check your assumption.
Common mistake Mistake 3: Confusing Q andK s p K_{sp} K s p Wrong approach: Calculating K s p K_{sp} K s p K s p K_{sp} K s p Why it feels right: You see concentrations, think "this is an equilibrium constant."
The fix: K s p K_{sp} K s p constant (from tables, temperature-dependent). Q is the trial product you compute from current concentrations. If Q > K s p Q > K_{sp} Q > K s p until Q Q Q K s p K_{sp} K s p
Steel-man: The notation is similar, and both involve ion concentrations raised to powers. The conceptual fix: K s p K_{sp} K s p
Common mistake Mistake 4: Forgetting Dilution When Mixing SolutionsWrong approach: Mixing 50 mL of 0.002 M Ag⁺ and 50 mL of 0.003 M Cl⁻, using [ \ce A g + ] = 0.002 [\ce{Ag+}] = 0.002 [ \ce A g + ] = 0.002
Why it feels right: You focus on the reaction stoichiometry and forget the volume change.
The fix: After mixing, total volume = 100 mL. Each concentration is halved:
[ \ce A g + ] = 0.002 × 50 100 = 0.001 M [\ce{Ag+}] = 0.002 \times \frac{50}{100} = 0.001 \text{ M} [ \ce A g + ] = 0.002 × 100 50 = 0.001 M [ \ce C l − ] = 0.003 × 50 100 = 0.0015 M [\ce{Cl-}] = 0.003 \times \frac{50}{100} = 0.0015 \text{ M} [ \ce C l − ] = 0.003 × 100 50 = 0.0015 M
Steel-man: In stoichiometry problems, you often ignore dilution (limiting reagent in moles doesn't depend on volume). Here, concentrations matter for equilibrium, so dilution is critical.
K s p K_{sp} K s p K s p K_{sp} K s p Δ G ∘ = − R T ln K s p \Delta G^\circ = -RT \ln K_{sp} Δ G ∘ = − R T ln K s p Δ G ∘ = Δ H ∘ − T Δ S ∘ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ Δ G ∘ = Δ H ∘ − T Δ S ∘ Δ H ∘ \Delta H^\circ Δ H ∘ Δ S ∘ \Delta S^\circ Δ S ∘ K s p K_{sp} K s p
Endothermic dissolution (Δ H ∘ > 0 \Delta H^\circ > 0 Δ H ∘ > 0 K s p K_{sp} K s p Exothermic dissolution (Δ H ∘ < 0 \Delta H^\circ < 0 Δ H ∘ < 0 Ca(OH) 2 \text{Ca(OH)}_2 Ca(OH) 2 K s p K_{sp} K s p
Mnemonic Remembering When to Precipitate"Q Quickly KSP"
Q > K : precipitation Q uickly happens.Q < K : K eeps dissolving (can take more).Q = K : equilibrium, s table/saturated (K SP).
Recall Feynman Explanation for a12-Year-OldImagine you're making lemonade. You keep adding sugar until no more dissolves—that's "saturated." The amount that dissolves is like solubility. Now, what if your lemonade already has a bunch of sugar in it from somewhere else (like you added pre-sweetened water)? If you try adding more sugar, way less will dissolve because the water is already "crowded" with sugar particles. That's the common-ion effect!
And let's say you have two types of sugar: one that dissolves super easily (like regular sugar) and one that barely dissolves (like rock candy). If you slowly add water, the rock candy will "fall out" (precipitate) first because it hits its limit soner. By controlling how much water you add, you can separate the two sugars. That's selective precipitation—fishing out the less-soluble stuff first!
The K s p K_{sp} K s p K s p K_{sp} K s p K s p K_{sp} K s p
Le Chatelier's Principle – explains shift in equilibrium when common ion is addedChemical Equilibrium andrium Constant – K s p K_{sp} K s p K c K_c K c Ionic Equilibria in Solutions – acids, bases, and bufers also involve equilibrium expressionsQualitative Inorganic Analysis – selective precipitation is the basis for separating cation groupsGibs Free Energy and Spontaneity – Δ G ∘ = − R T ln K s p \Delta G^\circ = -RT \ln K_{sp} Δ G ∘ = − R T ln K s p pH and pOH Calculations – controlling [ \ce S 2 − ] [\ce{S^{2-}}] [ \ce S 2 − ] [ OH − ] [\text{OH}^-] [ OH − ] Stoichiometry and Solution Concentration – dilution and mole calculations underpin mixing problems
#flashcards/chemistry
What is the solubility product constant K s p K_{sp} K s p M x A y ( s ) ⟷ x M n + + y A m − M_xA_y(s) \longleftrightarrow xM^{n+} + yA^{m-} M x A y ( s ) ⟷ x M n + + y A m − K s p = [ M n + ] x [ A m − ] y K_{sp} = [\text{M}^{n+}]^x [\text{A}^{m-}]^y K s p = [ M n + ] x [ A m − ] y
Why does adding a common ion suppress solubility? Le Chatelier: increasing a product ion shifts equilibrium toward reactants (solid). Mathematically, to keep
K s p K_{sp} K s p constant, if one ion's
[ ⋅ ] [\cdot] [ ⋅ ] rises, the other must fall—less dissolves.
For AgCl \text{AgCl} AgCl s ′ s' s ′ K s p = 1.8 × 10 − 10 K_{sp} = 1.8 \times 10^{-10} K s p = 1.8 × 1 0 − 10 Assume
[ Cl − ] ≈ 0.10 [\text{Cl}^-] \approx 0.10 [ Cl − ] ≈ 0.10 (common ion dominates). Then
K s p = s ′ ⋅ 0.10 ⟹ s ′ = K s p / 0.10 = 1.8 × 10 − 9 K_{sp} = s' \cdot 0.10 \implies s' = K_{sp}/0.10 = 1.8 \times 10^{-9} K s p = s ′ ⋅ 0.10 ⟹ s ′ = K s p /0.10 = 1.8 × 1 0 − 9 M. Check
s ′ ≪ 0.10 s' \ll 0.10 s ′ ≪ 0.10 afterward.
What is the ionic product Q and how does it predict precipitation? Q is the product of current ion concentrations (raised to stoichiometric powers), not at equilibrium. If
Q > K s p Q > K_{sp} Q > K s p , precipitation occurs; if
Q < K s p Q < K_{sp} Q < K s p , more can dissolve; if
Q = K s p Q = K_{sp} Q = K s p , saturated equilibrium.
Mixing solutions:50 mL of 0.002 M Ag⁺ + 50 mL of 0.004 M Cl⁻. What are the concentrations after mixing? Total volume 100 mL.
[ Ag + ] = 0.002 × ( 50 / 100 ) = 0.001 [\text{Ag}^+] = 0.002 \times (50/100) = 0.001 [ Ag + ] = 0.002 × ( 50/100 ) = 0.001 M.
[ Cl − ] = 0.004 × ( 50 / 100 ) = 0.002 [\text{Cl}^-] = 0.004 \times (50/100) = 0.002 [ Cl − ] = 0.004 × ( 50/100 ) = 0.002 M. Always account for dilution.
For CaF 2 \text{CaF}_2 CaF 2 K s p = 3.9 × 10 − 11 K_{sp} = 3.9 \times 10^{-11} K s p = 3.9 × 1 0 − 11 C a F 2 ⇌ C a 2 + + 2 F − CaF_2 \rightleftharpoons Ca^{2+} + 2F^- C a F 2 ⇌ C a 2 + + 2 F − . If solubility
= s = s = s , then
[ Ca 2 + ] = s [\text{Ca}^{2+}] = s [ Ca 2 + ] = s ,
[ F − ] = 2 s [\text{F}^-] = 2s [ F − ] = 2 s .
K s p = s ( 2 s ) 2 = 4 s 3 K_{sp} = s(2s)^2 = 4s^3 K s p = s ( 2 s ) 2 = 4 s 3 . Solve:
s = ( K s p / 4 ) 1 / 3 = ( 9.75 × 10 − 12 ) 1 / 3 ≈ 2.14 × 10 − 4 s = (K_{sp}/4)^{1/3} = (9.75 \times 10^{-12})^{1/3} \approx 2.14 \times 10^{-4} s = ( K s p /4 ) 1/3 = ( 9.75 × 1 0 − 12 ) 1/3 ≈ 2.14 × 1 0 − 4 M (so
[ F − ] = 2 s ≈ 4.3 × 10 − 4 [\text{F}^-] = 2s \approx 4.3 \times 10^{-4} [ F − ] = 2 s ≈ 4.3 × 1 0 − 4 M).
extra product ions shift back
Intuition Hinglish mein samjho
Intuition Hinglish mein samjhoDekho, is note ka core idea bahut simple hai — jab koi salt paani mein ghulta hai, toh ek equilibrium set hota hai jahan solid aur dissolved ions ke beech balance hota hai. Is balance ko hum ek number se measure karte hain jise K s p K_{sp} K s p K s p K_{sp} K s p
Ab asli mazedaar concept hai common-ion effect . Socho ek room already logon se bhara hua hai — ab naye log kam hi aa payenge, right? Waise hi, agar solution mein pehle se woh ion maujood hai jo salt bhi deta hai, toh Le Chatelier ke principle ke hisaab se equilibrium peeche shift ho jaata hai, aur salt ka aur zyada precipitate ho kar baith jaata hai. Matlab salt ki solubility kam ho jaati hai. Yeh samajhna zaroori hai kyunki solubility (s) aur K s p K_{sp} K s p K s p K_{sp} K s p AgCl \text{AgCl} AgCl K s p = s 2 K_{sp} = s^2 K s p = s 2 CaF 2 \text{CaF}_2 CaF 2 K s p = 4 s 3 K_{sp} = 4s^3 K s p = 4 s 3
Yeh cheez kyun matter karti hai? Kyunki isi principle se hum selective precipitation karte hain — yaani ek mixture mein se ions ko alag-alag precipitate kar ke separate karna. Yeh qualitative analysis (jo tumhare practicals mein aata hai) aur industry mein purification ka backbone hai. Toh yeh sirf theory nahi, real-world separation techniques ki foundation hai. Agar K s p K_{sp} K s p
