Question

1 litre of saturated solution of caco3 is evaporated to dryness then 7g of residue is left the solubility product of caco3 is

Answer 1

Answer : Solubility product of CaCO₃ is 4.89 x 10⁻³.

Explanation :

Step 1 : Find concentration of CaCO₃

Let us find the concentration of CaCO₃ first.

The concentration is calculated as mol/L.

Molar mass of CaCO3 is 100.1 g/mol

The moles of CaCO3 = $\frac{grams}{MolarMass}$

$Moles = \frac{7g}{100.1g/mol}=0.0699$

$Concentration = \frac{mol}{L}=\frac{0.0699mol}{1L}= 0.0699M$

Concentration of CaCO3 is 0.0699 M.

Step 2 : Find concentration of Ca and CO3 ions

The dissociation reaction of CaCO3 is written as,

$CaCO_{3} \rightarrow Ca^{2+}+CO_{3}^{2-}$

The mole ratio of CaCO3 and its ions is 1:1. Therefore concentration of ions is same as that of CaCO3.

[Ca²⁺] = 0.0699M

[CO3²⁻] = 0.0699M

Step 3 : Use solubility product formula

The solubility product of CaCO₃ can be written as,

$Ksp = [Ca^{2+}][CO_{3}^{2-}]$

$Ksp = (0.0699)(0.0699)$

Ksp = 0.00489

Solubility product of CaCO₃ is 4.89 x 10⁻³.

Answer 2

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