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 pK value for the formation of \(\mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}\) is \(-8.60 .\) The reaction is $$ \mathrm{Mg}^{2+}(a q)+\mathrm{P}_{3} \mathrm{O}_{10}^{5-}(a q) \rightleftharpoons \mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}(a q) $$ Calculate the concentration of \(\mathrm{Mg}^{2+}\) in a solution that was originally $50 . \mathrm{ppm} \mathrm{Mg}^{2+}(50 . \mathrm{mg} / \mathrm{L} \text { of solution) after } 40 . \mathrm{g} \text { of }\( \)\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\( is added to 1.0 \)\mathrm{L}$ of the solution.

Step by step solution

01

Calculate the concentration of sodium tripolyphosphate

First, we need to calculate the concentration of sodium tripolyphosphate (Na5P3O10) added to the solution. We are given that 40 g of Na5P3O10 is added to 1.0 L of the solution. To find the concentration in mol/L, we will divide the mass by the molar mass and the volume: Molar mass of Na5P3O10 = (5 × 22.99) + (3 × 30.97) + (10 × 16.00) g/mol = 367.87 g/mol Concentration of Na5P3O10 = (40 g) / (367.87 g/mol × 1.0 L) = 0.1087 mol/L

02

Calculate the reaction quotient, Q

Now we will use the given pK value (-8.60) to find the equilibrium constant, K: K = 10^(-pK) = 10^8.60 We are also given the initial concentration of Mg^2+ ions as 50 ppm, which is equal to 50 mg/L. To convert this to molarity, we need to divide by the molar mass of Mg^2+ (24.31 g/mol): Initial concentration of Mg^2+ = (50 mg/L) × (1 g / 1000 mg) × (1 mol / 24.31 g) = 0.002055 mol/L Next, we will calculate the reaction quotient, Q, using the concentrations of Mg^2+ and Na5P3O10: Q = [MgP3O10^(3-)] / ([Mg^2+] × [P3O10^(5-)]) At the start, [MgP3O10^(3-)] = 0, so Q = 0.

03

Determine the direction of the reaction

As Q < K, the reaction will move towards the right side (forward direction).

04

Calculate the final concentration of Mg^2+ ions

Let x be the moles of Mg^2+ that reacted. Thus, the final concentrations at equilibrium will be: [Mg^2+] = 0.002055 - x [P3O10^(5-)] = 0.1087 - x [MgP3O10^(3-)] = x We can now set up an expression for K using these concentrations: K = [MgP3O10^(3-)] / ([Mg^2+] × [P3O10^(5-)]) = x / ((0.002055 - x) × (0.1087 - x)) Since K ≫ 1, the reaction proceeds almost to completion, and we can assume that x ≈ 0.002055 mol/L. Therefore, the final concentration of Mg^2+ ions in the solution is: [Mg^2+] ≈ 0.002055 - x ≈ 0.002055 - 0.002055 ≈ 0 mol/L However, this is an approximation, so the final concentration of Mg^2+ ions will be very close to 0 mol/L, but not exactly 0.

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