Question

Solubility of \(Ca_3(PO_4)_2\) in \(100\) \(mL\) of pure water is \(W\) gm. Find out \(K_{sp}\) of \(Ca_3(PO_4)_2\) is: (M: Molecular mass of \(Ca_3(PO_4)_2\))

• A. \(108 \times \bigg(\frac{W}{M}\bigg)^5\)
• B. \(108 \times 105 \times \bigg[\frac{W}{M}\bigg]^5\)
• C. \(108 \times 104 \times \bigg[\frac{W}{M}\bigg]^5\)
• D. \(108 \times 106 \times \bigg[\frac{W}{M}\bigg]^5\)

Answer 1

The Correct Option is B

To find the solubility product constant, \(K_{sp}\), of \(Ca_3(PO_4)_2\), we start by writing its dissociation in water:

Ca_3(PO_4)_2 \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}

Let the solubility of \(Ca_3(PO_4)_2\) in water be \(s\) mol/L. Therefore, in a saturated solution:

• The concentration of \(Ca^{2+}\) ions = \(3s\) mol/L
• The concentration of \(PO_4^{3-}\) ions = \(2s\) mol/L

The expression for the solubility product \(K_{sp}\) is given by:

K_{sp} = [Ca^{2+}]^3 \cdot [PO_4^{3-}]^2

Substitute the ion concentrations into this expression:

K_{sp} = (3s)^3 \cdot (2s)^2

Calculate the above expression:

K_{sp} = 27s^3 \cdot 4s^2 = 108s^5

Now, convert the solubility from grams to moles. Given that \(W\) gm of \(Ca_3(PO_4)_2\) dissolves in 100 mL (0.1 L) of water:

• Molecular mass of \(Ca_3(PO_4)_2\) is \(M\) g/mol.
• So, \(s\) in moles/L is given by: s = \frac{W}{M \times 0.1} = \frac{10W}{M}

Substitute this into the \(K_{sp}\) expression:

K_{sp} = 108 \cdot \left(\frac{10W}{M}\right)^5

Simplify the expression:

K_{sp} = 108 \cdot 10^5 \cdot \left(\frac{W}{M}\right)^5 = 108 \times 100000 \times \left(\frac{W}{M}\right)^5

Therefore, the \(K_{sp}\) for \(Ca_3(PO_4)_2\) is:

K_{sp} = 108 \times 10^5 \times \left(\frac{W}{M}\right)^5

This matches with the correct option: 108 \times 105 \times \left[\frac{W}{M}\right]^5