The mass of CaC2O4 (in g) to be dissolved in distilled water to make 1.0 L of saturated solution is (KSP of CaC2O4 = 2.5 x 10-9 mol2L-2) molar mass of CaC2O4 = 128 g mol-1

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q131 – AP-EAMCET 2026

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Home Chemistry EquilibriumQuestion #416

The mass of

C a C_{2} O_{4}

(in g) to be dissolved in distilled water to make 1.0 L of saturated solution is (

K_{S P}

C a C_{2} O_{4} = 2.5 \times 10^{- 9} {mol}^{2} L^{- 2}

) molar mass of

C a C_{2} O_{4} = 128 {g mol}^{- 1}

Clue 1 of 3: Solubility Product Constant

The solubility product constant ( $K_{S P}$ ) describes the equilibrium between a sparingly soluble ionic solid and its ions in a saturated solution. It is a measure of the compound's solubility.

🧠 Full Solution Path

Step 1: Write the Dissociation Equation and Ksp Expression✦ Active

Calcium oxalate ( $C a C_{2} O_{4}$ ) is a sparingly soluble salt. When it dissolves in water, it dissociates into calcium ions ( $C a^{2 +}$ ) and oxalate ions ( $C_{2} O_{4}^{2 -}$ ). The equilibrium can be represented as:

C a C_{2} O_{4} (s) ⇌ C a^{2 +} (a q) + C_{2} O_{4}^{2 -} (a q)

If $s$ represents the molar solubility of $C a C_{2} O_{4}$ (in mol/L), then at equilibrium, the concentration of $C a^{2 +}$ ions is $s$ and the concentration of $C_{2} O_{4}^{2 -}$ ions is also $s$ . The solubility product constant ( $K_{S P}$ ) expression is:

K_{S P} = [C a^{2 +}] [C_{2} O_{4}^{2 -}] = s \times s = s^{2}

💡 Teacher's Secret Hint

Remember that the solid reactant ( $C a C_{2} O_{4} (s)$ ) is not included in the $K_{S P}$ expression.

Step 2: Calculate the Molar Solubility (s)○ Expand

We are given the $K_{S P}$ value for $C a C_{2} O_{4}$ as $2.5 \times 10^{- 9} {mol}^{2} L^{- 2}$ . We can use this to calculate the molar solubility $s$ :

s^{2} = K_{S P}

s^{2} = 2.5 \times 10^{- 9}

s = \sqrt{2.5 \times 10^{- 9}} = \sqrt{25 \times 10^{- 10}}

s = 5 \times 10^{- 5} mol/L

💡 Teacher's Secret Hint

Be careful with scientific notation when taking square roots. It's often easier to adjust the exponent to be an even number before taking the root, e.g., $2.5 \times 10^{- 9} = 25 \times 10^{- 10}$ .

Step 3: Calculate the Mass of CaC2O4 Required○ Expand

We need to make 1.0 L of a saturated solution. The molar solubility $s$ represents the moles of $C a C_{2} O_{4}$ that dissolve per liter of solution. Therefore, for 1.0 L, the moles of $C a C_{2} O_{4}$ required are:

Moles of C a C_{2} O_{4} = s \times Volume = (5 \times 10^{- 5} mol/L) \times (1.0 L) = 5 \times 10^{- 5} mol

The molar mass of $C a C_{2} O_{4}$ is given as $128 {g mol}^{- 1}$ . Now, we can convert moles to mass:

Mass of C a C_{2} O_{4} = Moles \times Molar Mass

Mass of C a C_{2} O_{4} = (5 \times 10^{- 5} mol) \times (128 g/mol)

Mass of C a C_{2} O_{4} = 640 \times 10^{- 5} g

Mass of C a C_{2} O_{4} = 0.00640 g

💡 Teacher's Secret Hint

Always ensure your units cancel out correctly to arrive at the desired unit (grams in this case). Moles/L * L * g/mol = g.

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Chemistry Question 131 – AP-EAMCET 2026

Clue 1 of 3: Solubility Product Constant

🧠 Full Solution Path