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Chapter 17: Problem 42

A volume of \(75 \mathrm{~mL}\) of \(0.060 \mathrm{M} \mathrm{NaF}\) is mixed with \(25 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate the concentra- tions in the final solution of \(\mathrm{NO}_{3}^{-}, \mathrm{Na}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{F}^{-} \cdot\left(K_{\mathrm{sp}}\right.\) for \(\left.\mathrm{SrF}_{2}=2.0 \times 10^{-10} \mathrm{.}\right)\)

Short Answer

Expert verified
The final concentrations of the ions in the solution are: [NO3^-] = 0.075 M, [Na+] = 0.045 M, [Sr2+] = 2.5 * 10^-5 M, [F-] = 0.045 M.

Step by step solution

01

Calculation of initial concentrations

Begin by calculating the initial concentrations of the individual ions. Remember to calculate the final total volume first. The reaction form is: \nSr(NO3)2 \(\rightarrow\) Sr2+ + 2NO3-\nNaF \(\rightarrow\) Na+ + F-\nSo, the final total volume is 75 mL + 25 mL = 100 mL = 0.1 L. Then we calculate initial concentrations of the ions:\n\n[Sr2+] = (0.15 moles/liter)x(0.025 liters)/(0.1 liters) = 0.0375 M\n\n[NO3-] = 2x[Sr2+] = 2x0.0375 M = 0.075 M\n\n[Na+] = (0.060 moles/liter)x(0.075 liters)/(0.1 liters) = 0.045 M\n\n[F-] = [Na+] = 0.045 M
02

Determine if a precipitate forms

Next, we use the solubility product constant (Ksp) to check if any insoluble substance forms from the reaction. In this case, we're concerned about strontium fluoride (SrF2). The solubility product constant (Ksp) expression is defined as: \nKsp = [Sr2+][F-]^2\nAfter substitution, we have: \n2 * 10^-10 = (0.0375 M) * (0.045 M)^2 = 7.59 * 10^-5\nThis value is greater than the given Ksp, therefore, a precipitate of SrF2 will form.
03

Calculation of equilibrium concentrations

Since we know that a precipitate forms, the concentrations of Sr2+ and F- will change until reaching an equilibrium point with the formation of SrF2. Let x be the decrease in the concentrations of Sr2+ and F- due to the formation of SrF2. Then, the new concentrations will be:\n[Sr2+] = 0.0375 - x\n[F-] = 0.045 - 2x\nSubstituting these into the Ksp expression, we solve for x. We get x = 0.0375 M (as the decrease in the concentration of Sr2+ is almost complete) assuming that the fluoride ion is in excess.
04

Computing final concentrations

By knowing the decrease in concentrations of Sr2+ and F- (x), we can now calculate the final concentrations. Substitute the solved x value into the expressions of [Sr2+] and [F-] and find the final equilibrium concentration. Therefore:\n[NO3^-] = 0.075 M (since the nitrate ion is a spectator ion, its concentration does not change)\n[Na+] = 0.045 M (since the sodium ion is a spectator ion, its concentration does not change)\n[Sr2+] = 0.0375 M - x = 2.5 * 10^-5 M (this is the saturation concentration) \n[F-] = 0.045 M - 2x = 0.045 M (as the decreased concentration of fluoride ion can be neglected)

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Most popular questions from this chapter

What is the \(\mathrm{pH}\) of the buffer \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4} / 0.15 \mathrm{M}\) \(\mathrm{KH}_{2} \mathrm{PO}_{4} ?\)

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