Calculate the solubility of silver phosphate, Ag3PO3 –>3Ag+ + PO33- , in pure water.
Ksp = 2.6 x 10-18 correct answer: 1.8 x 10^-5 M please show me step by step, and as detailed as possible, how to solve this problem. the correct answer is already written above ,thank you so much 🙂
The Correct Answer and Explanation is :
To calculate the solubility of silver phosphate (Ag₃PO₄) in pure water, we will use its solubility product constant ((K_{sp})) and apply the equilibrium concepts.
Step-by-Step Solution:
- Write the dissociation equation:
Silver phosphate dissociates in water as follows: [
Ag_3PO_4(s) \rightleftharpoons 3Ag^+(aq) + PO_4^{3-}(aq)
] From the dissociation equation, you can see that for every 1 mole of silver phosphate that dissolves, 3 moles of silver ions (Ag⁺) and 1 mole of phosphate ions (PO₄³⁻) are produced. - Set up an ICE (Initial, Change, Equilibrium) table:
Let ( s ) be the molar solubility of Ag₃PO₄ in pure water, which is the amount that dissolves to form Ag⁺ and PO₄³⁻. Ion Initial Concentration (M) Change (M) Equilibrium Concentration (M) (Ag^+) 0 +3s 3s (PO_4^{3-}) 0 +s s - Write the expression for the solubility product, (K_{sp}): The (K_{sp}) expression for silver phosphate is: [
K_{sp} = [Ag^+]^3 [PO_4^{3-}]
] Substituting the equilibrium concentrations: [
K_{sp} = (3s)^3 \cdot (s)
] This simplifies to: [
K_{sp} = 27s^4
] - Substitute the given (K_{sp}) value: We are given (K_{sp} = 2.6 \times 10^{-18}), so: [
2.6 \times 10^{-18} = 27s^4
] - Solve for (s): First, divide both sides of the equation by 27: [
s^4 = \frac{2.6 \times 10^{-18}}{27} = 9.63 \times 10^{-20}
] Next, take the fourth root of both sides: [
s = \sqrt[4]{9.63 \times 10^{-20}}
] Using a calculator: [
s \approx 1.8 \times 10^{-5} \, \text{M}
]
Thus, the solubility of silver phosphate in pure water is approximately (1.8 \times 10^{-5} \, \text{M}).
Explanation:
The key to solving this problem is recognizing that the solubility product constant ((K_{sp})) relates the concentrations of the ions in equilibrium. By setting up an ICE table and substituting the equilibrium concentrations into the (K_{sp}) expression, we can solve for the solubility of Ag₃PO₄ in water.
The solubility constant is very small ((K_{sp} = 2.6 \times 10^{-18})), indicating that silver phosphate is only sparingly soluble in water. This low solubility means that only a small amount of Ag₃PO₄ dissolves to form ions in solution.
The solubility result, (1.8 \times 10^{-5} \, \text{M}), tells us how much silver phosphate can dissolve to reach equilibrium at a given temperature.