Question

Sodium tripolyphosphate \(\left(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\right)\) is used in many synthetic detergents. Its major effect is to soften the water by complexing \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) ions. It also increases the efficiency of surfactants, or wetting agents that lower a liquid's surface tension. The \(K\) value for the formation of \(\mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}\) is \(4.0 \times 10^{8} .\) The reaction is \(\mathrm{Mg}^{2+}+\mathrm{P}_{3} \mathrm{O}_{10}^{5-} \rightleftharpoons \mathrm{MgP}_{3} \mathrm{O}_{10}{ }^{3-} .\) Calculate the concentration of \(\mathrm{Mg}^{2+}\) in a solution that was originally \(50 . \mathrm{ppm} \mathrm{Mg}^{2+}(50 . \mathrm{mg} / \mathrm{L}\) of solution) after 40. g \(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\) is added to \(1.0 \mathrm{~L}\) of the solution.

Short answer

To find the final concentration of Mg²⁺ ions after adding 40 g of sodium tripolyphosphate to the solution, first calculate the initial moles of Mg²⁺ and sodium tripolyphosphate, and then set up the ICE (Initial, Change, Equilibrium) table. Next, use the equilibrium constant value to find the value of \(x\), the change in moles. Finally, calculate the equilibrium moles of Mg²⁺ and divide by the volume of the solution (1.0 L) to obtain the final concentration of \(\mathrm{Mg}^{2+}\). Note that solving the equilibrium equation may require simplifying assumptions or numerical methods, such as graphical methods, Newton-Raphson method, or solver functions in calculators or spreadsheet software.

Step-by-step solution

Calculate the initial moles of Mg²⁺ and sodium tripolyphosphate

First, we need to find the initial moles of the \(\mathrm{Mg}^{2+}\) ions and sodium tripolyphosphate. The initial concentration of \(\mathrm{Mg}^{2+}\) is provided as 50 ppm (50 mg/L). To find the moles, we'll use the molar mass of Mg²⁺, which is 24.3 g/mol: \[\text{moles } \mathrm{Mg}^{2+} = \frac{50 \times 10^{-3}\, \text{g/L}}{24.3\, \text{g/mol}}\] Now, for sodium tripolyphosphate, we have 40 g added to 1.0 L of the solution. The molar mass of \(\mathrm{Na}_{5} \mathrm{P}_{3}\mathrm{O}_{10}\) is 367.9 g/mol: \[\text{moles } \mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10} = \frac{40 \, \text{g}}{367.9\, \text{g/mol}} \]

Set up the ICE table

Next, we'll set up an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations of the ions involved in the reaction. The initial concentration of \(\mathrm{P}_{3} \mathrm{O}_{10}^{5-}\) ions is zero since only \(\mathrm{Mg}^{2+}\) ions are present initially. | | Mg²⁺ | P₃O₁₀⁻⁵ | MgP₃O₁₀⁻³ | |-----------------|------|---------|-----------| | Initial (moles) | a | 0 | 0 | | Change (moles) | -x | +x | -x | | Equilibrium | a-x | x | x | #aequation#Where x is the change in moles, a is the initial moles of Mg²⁺, and x is the moles of P₃O₁₀⁻⁵ and MgP₃O₁₀⁻³ formed at equilibrium. The equilibrium constant expression for this reaction is: \[ K = \frac{[\mathrm{MgP}_{3} \mathrm{O}_{10}{ }^{3-}]}{[\mathrm{Mg}^{2+}][\mathrm{P}_{3} \mathrm{O}_{10}^{5-}]} \] Since we have the equilibrium constant value (\(K=4.0 \times 10^8\)), we can substitute the values from the ICE table into the equation: \[ 4.0 \times 10^8 = \frac{x^2}{(a-x)\cdot x} \]

Calculate the final concentration of Mg²⁺

To find the final concentration of Mg²⁺ ions, we first need to solve the equation above to get the value of \(x\). Do note, however, that solving this expression analytically might be complex. Therefore, it is advisable to either make the reasonable assumption that \(x \ll a\) and simplify the expression or use numerical methods (e.g., a graphical method, Newton-Raphson method, solver functions in calculators or spreadsheet software) for solving the equation. Once the value of \(x\) is found, use it to calculate the equilibrium moles of \(\mathrm{Mg}^{2+}\): \[\text{moles } \mathrm{Mg}^{2+} = a - x\] Then, divide by the volume of the solution (1.0 L) to obtain the final concentration of Mg²⁺: \[\text{Concentration of }\mathrm{Mg}^{2+} = \frac{\text{moles }\mathrm{Mg}^{2+}}{\text{Volume of solution}}\]

Key concepts

Chemical Equilibrium

When a reaction occurs in a closed system and reaches a state where the rate of the forward reaction is equal to the rate of the backward reaction, it is said to have achieved chemical equilibrium. This does not imply that the reactants and products are present in equal amounts, but rather that their concentrations have stabilized and will not change without an external influence (like a change in temperature, pressure, or the addition of a catalyst).

In the example given with sodium tripolyphosphate, when it is added to water containing magnesium ions, the resulting equilibrium between the magnesium ions (Mg²⁺), the tripolyphosphate ions (P₃O₁₀^5-), and the magnesium tripolyphosphate complex (MgP₃O₁₀^3-) has profound implications for the solubility and the eventual concentration of magnesium ions in the solution. Students should understand that at equilibrium, the amount of Mg²⁺ ions being complexed is equal to the amount of the MgP₃O₁₀^3- complex dissociating back into Mg²⁺ and P₃O₁₀^5- ions.

Complexation Reactions

Complexation reactions are reactions where simple ions come together to form a more complex structure known as a coordination complex. The complex often involves a central metal ion (such as Mg²⁺) bonded to one or more molecules or ions (ligands). These can significantly impact the properties of the solution, including solubility, color, and reactivity.

In the detergent case involving sodium tripolyphosphate, the purpose of this complexing agent is to bind calcium and magnesium ions that cause water hardness. The complexation reaction is so effective that it can drastically reduce the 'free', uncomplexed magnesium ions in the solution, which is crucial in allowing detergents to act more efficiently. Students should recognize that complexation directly influences the apparent solubility of ions in the water and the effectiveness of agents like tripolyphosphate.

Solubility Product Constant

The solubility product constant (Ksp) is a specialized equilibrium constant for the solubility of a compound. It is the product of the concentrations of the ions in a saturated solution raised to the power of their coefficients in the equilibrium equation. For slightly soluble salts, the Ksp is a measure of how much the salt can dissociate into its constituent ions in solution.

In the exercise, while the Ksp isn't provided or asked for directly, understanding this concept can help students appreciate the underlying principles governing the equilibrium. The larger the value of the complex’s equilibrium constant (as given K = 4.0 x 10⁸ for the formation of MgP₃O₁₀^3-), the more favorable the complex formation is. Essentially, a high K value indicates a very low concentration of uncomplexed ions, implying that the process of complexation is very effective. This helps explain why, under certain conditions, a solution can appear to hold more of a substance than its Ksp would suggest, especially in the presence of complexing agents like tripolyphosphate.