Solubility Product (Kₛₚ) , Solubility & Precipitation Numericals Made Super Easy | The Ultimate MDCAT Chemistry Numericals | Learn Chemistry by Inam Jazbi

Welcome to Learn Chemistry by Inam Jazbi, the platform where complex chemistry concepts become super-easy. Today’s post covers one of the most high-yield MDCAT topics—Solubility Product (Kₛₚ), Solubility, Ionic Equilibria, and Precipitation Numericals. If you want to master Kₛₚ tricks, shortcut methods, exam-focused questions, and the logic behind precipitation, this guide is made exactly for you.

Students often get confused when to precipitate, how to compare Ksp values, and how to solve solubility numericals quickly. In this blog, everything is explained in a simple, colourful, exam-oriented style so you can score full marks in MDCAT and board exams.

Let’s start mastering Kₛₚ — the smart way!

Q1. Write the solubility product expression for the following slightly soluble salts; CaSO₄, PbCl₂ Fe(OH)₃, BaF₂, Li₂C₂O₄, MgCO₃, Ag₃PO₄.                                                        

Solution

To write Kₛₚ expression, first of all write the balanced equation of ionic equilibrium of given compound to find out coefficient of each ion. Then Kₛₚ expression is written as product of ionic concentrations of cation and anion of given compound enclosed in square bracket with each concertation is raised to power equal to the coefficient of respective ion.

CaSO_4(s) ⇌ Ca²⁺_(aq) + SO₄²⁻_(aq)     Kₛₚ = [Ca²⁺] [SO₄²⁻]

PbCl_2(s) ⇌ Pb²⁺_(aq) + 2Cl⁻_(aq)      Kₛₚ = [Pb²⁺] [Cl⁻]²

Fe(OH)_3(s) ⇌ Fe³⁺_(aq) + 3OH⁻_(aq) Kₛₚ = [Fe³⁺] [OH⁻]³

Q. Write down the solubility product expressions for the following sparingly soluble salts along with their units

(i) BaF₂             

(ii) Li₂C₂O₄        

(iii) MgCO₃  

(iv) Ag₃PO₃      

(v) CaSO₄

(vi) AlF₃         

(vii) MgCr₂O₇   

(viii) PbI₂ 

(ix) Mg(OH)₂   

(x) Mg₃(PO₄)₂             

(xi) PbCrO₄ 

Answer

Q. Write the equation for dissociation of following salts and write down Kₛₚ expression.

(i) PbBr₂            (ii) Hg₂CrO₄   (iii) BaC₂O₄                   (iv) Fe(OH)₃

(v) Ag₂CO₃     (vi) Sb₂S₃         (vii) AgCNS                   (viii) Ag₃PO₄

(ix) Li₃Na₃(AlF₆)₂        (x) Hg₂I₂          

(xi) Ba₃(PO₄)₂             (xii) Ca₅(PO₄)₃F

(xiii) A₃B₄                    (xiv) CaF₂        

(xv) Ag₂CrO₄

Answer

Calculating Kₛₚ from given solubility Step-by-Step

Step 1 — Write the dissociation equation.

Write the solid ⇌ ions showing stoichiometric coefficients.

Step 2 — Define solubility, s.

Let s = molar solubility (mol L⁻¹) = moles of formula units dissolved per litre.

Step 3 — Express ion concentrations in terms of s.

From the stoichiometry, write each ion concentration as a function of s (and of any fixed/common ions present).

Step 4 — Write the Kₛₚ expression.

Multiply the equilibrium concentrations raised to their powers (from the balanced dissociation).

Step 5 — Substitute concentrations and solve for Ksp (or solve for s if Ksp is given).

Do algebra to get Ksp in numeric form. If solubility is given in g per volume convert to mol L⁻¹ first:

mol L⁻¹ = (mass in g per L) / (molar mass in g mol⁻¹)

Step 6 — Watch units and significant figures.

Report Kₛₚ as a dimensionless number (concentration units cancel out) using appropriate significant figures.

---

Common patterns (remember these!)

MX ⇌ M⁺ + X⁻ → [M] = s, [X] = s → Kₛₚ = s²
MX₂ ⇌ M²⁺ + 2X⁻ → [M] = s, [X] = 2s → Kₛₚ = [M][X]² = s(2s)² = 4 s³
M₂X₃ ⇌ 2M³⁺ + 3X²⁻ → [M] = 2s, [X] = 3s → Kₛₚ = (2s)²(3s)³ = 4·27·s⁵ = 108 s⁵

(Use the stoichiometric coefficients to build the expression.)

Q2.   Silver sulphate (Ag₂SO₄) is used for medicinal purpose to fill wounds. Its solubility in water at 25^oC  is 1.43 × 10^–2 mol/dm³. What will be its K_sp? (Example 7.8; Page 157)

Solution

The net ionic equation for ionization of Ag₂SO₄ is:

Ag₂SO_4(s)  ⇌   2Ag⁺_(aq)  +    SO₄^2–_(aq)

1 mol               2mol                1mol

x M                  2x M                  x M

Solubility of Ag₂SO₄ = x =  1.05 × 10^–5 mol/dm³

[Ag⁺]    = 2x = 2 × 1.43 x 10^–2 = 2.86 × 10^–2 mol/dm³

[SO₄^2–] =  x   = 1.43 × 10^–2 mol/dm³

Kₛₚ of Ag₂SO₄ =  ?

The Kₛₚ expression is:

Kₛₚ  = [Ag⁺]² [SO₄^2–]  

Kₛₚ  =   (2.86 × 10^–2 mol/dm³)²  x  (1.43 × 10^–2 mol/dm³)

Kₛₚ  =    1.17 × 10^–5 mol³/dm⁹

Q3. The Kₛₚ of Zn(OH)₂ is 2.1 × 10^–16 mol³/dm⁹ at 25C. Calculate its solubility in g/dm³ (The atomic mass of Zn = 65.4).   

 Solution

Kₛₚ of Zn(OH)₂       = 2.1 x 10^–16 mol³/dm⁹

Let,

Solubility of Zn(OH)₂ = x mol/dm³

The ionization equation of Zn(OH)₂  is:

Zn(OH)₂(s) ⇌ Zn²⁺                 + 2 OH⁻
x mol/dm³      x mol/dm³         2x mol/dm³

Solubility in mol/dm³

The Kₛₚ expression is:

Kₛₚ = [Zn²⁺]  [OH^–]² ⇒  2.1 x 10^–16 = (x). (2x)²  ⇒ 2.1 x 10^–16 =    (x).(4x²)  ⇒   4x³ = 2.1 x 10^–16

Solubility in g/dm³

Solubility of Zn(OH)₂ in mol/dm³           = 3.74 × 10⁻⁶ mol/dm³

Molecular mass of Zn(OH)₂ = 65.4 + 2(16) + 2(1) = 99.4 g/mol

Solubility of Zn(OH)₂ in g/dm³  = Solubility in mol/dm³   ×   Molar mass (molecular mass)

Solubility of Zn(OH)₂ in g/dm³  = 3.74 × 10⁻⁶ × 99.4 

Solubility of Zn(OH)₂ in g/dm³ = 3.717 × 10⁻⁴ g/dm³

Q4.Lead fluoride (PbF₂) is a high melting white solid used in glass coating to reflect IR rays. Its solubility in water at 25℃ is 0.58 g/dm³. Calculate its K_sp. (Atomic mass of Pb = 207 and F = 19) 5.3 × 10⁻⁸ mol³/dm⁹

Solution

Solubility of PbF₂ = 0.58 g/dm³

Molar mass of PbF₂ =  207 +  2(19) = 245g/mol

Solubility of PbF₂ in mol/dm³ = x = Solubility in g/dm³/ molar mass  =0.58 /245  = 2.37 × 10^–3 mol/dm³

The net ionic equation for ionization of Ag₂SO₄ is:

PbF_2(s)  ⇌   Pb²⁺_(aq)  +  2F^–_(aq)

1 mol            1mol             2mol

x M                 x M              2x M

Solubility of PbF₂ = x = 2.37 × 10^–3 mol/dm³

[Pb²⁺] = x = 2.37 ×10^–3 mol/dm³

[F^–] =  2x   = 2 × 2.37 × 10^–3 mol/dm³ = 4.74 × 10⁻³ mol/dm³

Kₛₚ of PbF₂ =  ?

The Kₛₚ expression is:

Kₛₚ  = [Pb²⁺] [F^–]²  = (2.37 × 10^–3 mol/dm³)  ×  (4.74 × 10⁻³ mol/dm³)² =  (2.37 × 10^–3) × (2.25×10⁻⁵) mol³/dm⁹

Kₛₚ  =    5.33 × 10^–5 mol³/dm⁹

Q1. Calculate Kₛₚ of BaSO₄ whose solubility is 1.05 × 10^–5 mole/dm³.

Solution

The net ionic equation for ionization of BaSO₄ is:

BaSO_{4 (s)}  ⇌   Ba²⁺_(aq)  +  SO₄^2–_(aq)

1 mol              1mol            1mol

x M                 x M                x M

Solubility of BaSO₄ = x =  1.05 × 10^–5 mol/dm³

[Ba²⁺]   = x = 1.05 × 10^–5 mol/dm³

[SO₄^2–] =  x = 11.05 × 10^–5 mol/dm³

Kₛₚ of BaSO₄ =  ?

The Kₛₚ expression is:

Kₛₚ  = [Ag⁺]² [SO₄^2–]  =   (1.05 × 10^–5 mol/dm³)  ×  (1.05 × 10^–5 mol/dm³)

Kₛₚ  =  1.1 x 10^–10 mol²/dm⁶

Q2. Calculate Kₛₚ of BaF₂ when its solubility is 0.0065 M at 25°C.

Solution

The ionization equation of BaF₂ is:

BaF₂                      ⇌           Ba²⁺                      +             2F^–

x M                                      x M                                      2x M

0.0065 M            0.0065 M            2(0.0065) M

Solubility of BaF₂ = x =  0.0065 M

[Ba²⁺]  = x = 1.05 x 10^–5 M

[F^–]   =  2x = 2 × 0.0065  M

Kₛₚ of BaF₂ = ?

The Kₛₚ expression is:

Kₛₚ    = [Ba²⁺]  [F^–]²        

Kₛₚ    = (0.0065)  (2 × 0.0065)²  = 0.0065  ×  (0.013)²     = 0.0065  ×  0.000169 = 0.00000109

Kₛₚ = 1.09 × 10^–6 M³

Q3. Calculate Kₛₚ of AgCl at 25°C when its solubility is 1.4 x 10^–3 g/dm³.

Solution

Solubility of AgCl = 1.4 × 10^–3 g/dm³

Molar mass of AgCl   =  108  +  35.5 = 143.5 g

Solubility of AgCl in mol/dm³ = Solubility in g dm^–3/ molar mass  =1.4 × 10^–3 /143.5  = 9.75 × 10^–6 M

The ionization equation of AgCl is:

AgCl                         ⇌   Ag⁺                +      Cl^–

x M                                 x M                         x M

9.75 × 10^–6 M         9.75 × 10^–6 M    9.75 × 10^–6 M

Solubility of AgCl = x =  9.75 × 10^–6 M

[Ag⁺]  = x = 9.75 × 10^–6 M

[Cl^–] = x = 9.75 × 10^–6 M

K_SP of AgCl = ?

The Kₛₚ expression is:

Kₛₚ = [Ag⁺] [Cl^–] = 9.75 × 10^–6 M × 9.75 × 10^–6 M² 

Kₛₚ  = 9.5 × 10^–11 M²

Calculating Solubility from given K_SP

Q1. Find the solubility of PbCrO₄ when its Kₛₚ is 2.8 × 10^–13 mol²/dm⁶.

Solution

Kₛₚ of PbCrO₄     = 2.8 x 10^–13 mol²/dm⁶

Let,

Solubility of PbCrO₄= x mol/dm³

The ionization equation of PbCrO₄ is:

PbCrO₄      ⇌           Pb²⁺                      +      CrO₄²⁻

x mol/dm³            x mol/dm³                   x mol/dm³

Solubility in mol/dm³

The Kₛₚ expression is:

Kₛₚ = [Pb²⁺]  [CrO₄²⁻]  ⇒   2.8 x 10^–13 = (x).(x)    ⇒   x²     = 2.8 x 10^–13 ⇒   x = √2.8 x 10^–13

x =   5.29 × 10^–7 mol/dm³

Solubility in g/dm³

Solubility of PbCrO₄ in mol/dm³= 5.29 × 10^–7 mol/dm³

Molecular mass of PbCrO₄ = 207 + 52 + 4(16) = 323 g/mol

Solubility of PbCrO₄ in g/dm³ = Solubility in mol/dm³   ×   Molar mass (molecular mass)

Solubility of PbCrO₄ in g/dm³    = 5.29 × 10^–7  ×  323 = 1.70 × 10^–4 g/dm³

Q2. Find the solubility of MgF₂ when its Kₛₚ is 7.26 × 10^–9 mol³/dm⁹.

Solution

Kₛₚ of MgF₂        = 7.26 × 10^–9 mol³/dm⁹

Let,

Solubility of MgF₂ = x mol/dm³

The ionization equation of MgF₂ is:

MgF₂            ⇌           Mg²⁺        +             2F^–

x mol/dm³           x mol/dm³             2x mol/dm³

The Kₛₚ expression is:

Kₛₚ = [Mg²⁺]  [F^–]² ⇒  7.26 × 10^–9 = (x). (2x)²  ⇒ 7.26 × 10^–9 = (x).(4x²)  ⇒   4x³ = 7.26  ×  10^–9

x³ = 7.26  × 10^–9/4  ⇒  x³ = 1.815  ×  10^–9 ⇒   x = ∛1.815  ×  10^–9  

x  = 1.219 × 10^–3 mol/dm³

Here,

Solubility of MgF₂ in g/dm³= 1.219 x 10^–3 mol/dm³

Molecular weight of MgF₂  = 24 + 38 = 62 g

Solubility of MgF₂ in g/dm³ = Solubility in mol/dm³   x   Molar mass (molecular mass)

Solubility of MgF₂ in g/dm³ = 1.219 × 10^–3  x  62 = 7.55 × 10^–2 g/dm³

Q3.   What is the ionic concentration of Ag⁺ and CrO₄²⁻ ions in a saturated solution of Ag₂CrO₄ at 25^oC? K_sp of Ag₂CrO₄ is 1.9  × 10^-12 mol³/dm⁹ (KB – 2015)

Solution

Kₛₚ of Ag₂CrO₄  = 1.9  x 10⁻¹² mol³/dm⁹

Let,

Solubility of Ag₂CrO₄  =  x mol/dm³

The ionization equation of Ag₂CrO₄  is:

Ag₂CrO₄        ⇌ 2Ag⁺                 + CrO₄²⁻
1                            2                          1
x mol/dm³       2x mol/dm³       x mol/dm³

The Kₛₚ expression is:

Kₛₚ = [Ag⁺]² [CrO₄^2–]  ⇒  1.9  x 10^–12  = (x).(2x)²  ⇒ 1.9  x 10^–12 = (x).(4x²)  ⇒   4x³ = 1.9  x 10^–12

x = 7.8 × 10⁻⁵  moldm⁻³

 

[Ag⁺]² = 2x = 2 × 7.8 × 10⁻⁵ = 1.56 × 10⁻⁴ moldm⁻³

[CrO₄²⁻] = x = 7.8 × 10⁻⁵ moldm⁻³

Q4.   If the concentration of CrO₄²⁻​ ions in a saturated solution of silver chromate is 2×10⁻⁴; calculate solubility product of silver chromate.

Solution

The ionization equation of Ag₂CrO₄ is:

Ag₂CrO₄               ⇌           2Ag⁺                 +             CrO₄^2–

1                                             2                                       1

x mol/dm³                 2x mol/dm³                      x mol/dm³

2×10⁻⁴ mol/dm³       2 (2×10⁻⁴)mol/dm³      2×10⁻⁴mol/dm³          

On the basis of this equation, the concentration of Ag⁺​ ion will be double of the concentration CrO₄^2– ions.

[Ag⁺]        = 2 × 2 ×10⁻⁴ M = 4 ×10⁻⁴ M

[CrO₄²⁻​] = 2 ×10⁻⁴ M

Kₛₚ ​= [Ag⁺]²[CrO₄²⁻​] = [4 ×10⁻⁴ M]² [2 ×10⁻⁴ M] = 3.2×10⁻¹¹ M³

Calculating precipitation occurs or not from Text Book

Calculating precipitation occurs or not from Past Papers

💥 غزل ۔۔۔۔ جونؔ ایلیا 💥

🌙 زندانیانِ شام و سحر خیریت سے ہیں
💭 ہر لمحہ جی رہے ہیں مگر خیریت سے ہیں

🏙️ شہرِ یقیں میں اب کوئی دم خم نہیں رہا
🌌 دشتِ گماں کے خاک بسر خیریت سے ہیں

🙏 آخر ہے کون جو کسی پل کہ سکے یہ بات
🕊️ اللہ اور تمام بشر خیریت سے ہیں

💡 ہے اپنے اپنے طور پہ ہر چیز اس گھڑی
💧 مژگانِ خشک و دامنِ تر خیریت سے ہیں

👀 اب فیصلوں کا کم نظروں پر مدار ہے
🌟 یعنی تمام اہلِ نظر خیریت سے ہیں

🦶 پیروں سے آبلوں کا وہی ہے معاملہ
🔥 سودائیانِ حال کے سر خیریت سے ہیں

🏚️ ہم جن گھروں کو چھوڑ کے آئے تھے ناگہاں
💭 شکوے کی بات ہے، وہ اگر خیریت سے ہیں

🌬️ لو چل رہی ہے، محو ہے اپنے میں دوپہر
🏚️ خاک اڑ رہی ہے اور کھنڈر خیریت سے ہیں

👥 ہم اہلِ شہر اپنے جوانوں کے درمیاں
✨ جونؔ! ایک معجزہ ہے اگر خیریت سے ہیں

🎨 برباد ہوچکا ہے ہنر اک ہنر کے ساتھ
📜 اور اپنے صاحبانِ ہنر خیریت سے ہیں

🕊️ شکرِ خدا شہید ہوئے اہلِ حق تمام
⚔️ برگستوان و تیغ و تبر خیریت سے ہیں

🏰 اب اس کا قصرِ ناز کہاں اور وہ کہاں
🚪 بس در ہے اور بندئہ در خیریت سے ہیں

✍️ ہم ہیں کہ شاعری ہے ہمارے لئے عذاب
📖 ورنہ تمام جوشؔ و جگرؔ خیریت سے ہیں

📚 شاعر تو دو ہیں میرؔ تقی اور میر جونؔ
🌙 باقی جو ہیں وہ شام و سحر خیریت سے ہیں

✍️ جونؔ ایلیا