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Determine the approximate solubility of \(\mathrm{AgCl}\) in \(0.10 \mathrm{M} \mathrm{NaCl}\) solution. \(\mathrm{K}_{\mathrm{sp}}\) for \(\mathrm{AgCl}=1.1 \times 10^{-10}\).

Short Answer

Expert verified
The approximate solubility of AgCl in a 0.10 M NaCl solution is \(1.1 \times 10^{-9}\) M.

Step by step solution

01

Write the chemical equilibrium expression for the dissolution of AgCl

For the dissolution of AgCl in water, we can write the following chemical equilibrium expression: \(AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}\)
02

Write the solubility product expression for AgCl

The solubility product (Ksp) is the product of the concentrations of the ions in a saturated solution of a salt at equilibrium, raised to the power of their stoichiometric coefficients. For AgCl, the Ksp expression is: \[K_{sp} = [Ag^+][Cl^-]\]
03

Consider the common ion effect

The common ion effect states that the solubility of a salt is lowered when it is dissolved in a solution that already contains an ion in common with the salt. In this problem, NaCl is the common ion source for Cl鈦. Since NaCl is a strong electrolyte and fully dissociates in water, the concentration of Cl鈦 ions is equal to the concentration of NaCl, which is 0.10 M.
04

Write the equilibrium expression for AgCl in the presence of NaCl

In the presence of NaCl, the equilibrium expression for AgCl becomes: \[K_{sp} = [Ag^+][(0.10 +x)]\] Here, x is the concentration of Ag鈦 ions resulting from the dissolution of AgCl, and 0.10 + x represents the total concentration of Cl鈦 ions (0.10 M contributed by NaCl and x contributed by AgCl).
05

Solve for x

Since the solubility of AgCl is very low (indicated by the small Ksp value), the x value will be much smaller than 0.10. Therefore, we can simplify the equilibrium expression by considering x to be negligible compared to 0.10, yielding: \[K_{sp} = [Ag^+][0.10]\] Now, solve for the concentration of Ag鈦 ions (x): \[x = \frac{K_{sp}}{0.10}\] Substitute the known values of Ksp into the equation: \[x = \frac{1.1 \times 10^{-10}}{0.10}\]
06

Calculate the solubility of AgCl

Calculate x, which is the solubility of AgCl in the 0.10 M NaCl solution: \[x = 1.1 \times 10^{-9} \mathrm{M}\] So, the approximate solubility of AgCl in 0.10 M NaCl solution is \(1.1 \times 10^{-9}\) M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant (Ksp)
Understanding the solubility product constant, or Ksp, is crucial when examining how salts dissolve in water. Ksp indicates how much of a salt can dissolve in a solution until it reaches a state of equilibrium. Consider it a guideline for how well a salt dissolves.

The Ksp value is characterized as the product of the concentrations of the ions in a saturated solution. Essentially, it tells us the maximum amount of dissolved ions a solution can handle at equilibrium. For silver chloride (AgCl), this expression is represented as \[K_{sp} = [Ag^+][Cl^-]\].

This equation means that when \([Ag^+]\) and \([Cl^-]\) ions are in balance in a solution saturated with AgCl, the product of their concentrations equals the Ksp value of AgCl鈥攌nown to be \(1.1 \times 10^{-10}\). This extremely low Ksp suggests AgCl is only sparingly soluble in water."},{"concept_headline":"Chemical Equilibrium","text":"In chemical systems, equilibrium is like a balance point. It's a state where the forward and reverse reactions occur at equal rates, so the concentrations of reactants and products remain constant over time.

For the salt AgCl, achieving chemical equilibrium in water means the rate at which AgCl dissolves to form ions is equal to the rate at which the ions recombine to form solid AgCl again. This interaction can be visualized as \[AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}\].

At equilibrium, this delicate balance ensures that a saturated solution of a salt contains constant concentrations of the ions involved, as dictated by the Ksp expression. Changes to the solution, such as adding more ions, can shift this equilibrium and affect solubility."},{"concept_headline":"Common Ion Effect","text":"The common ion effect is a fascinating occurrence that can greatly influence solubility. It revolves around the introduction of additional ions into an existing solution that already contains them, either from a dissolved salt or another source.

When a solution already has an ion that is part of the salt's solubility equilibrium, like in our example involving AgCl in NaCl solution, the common ion effect comes into play. Since NaCl dissociates completely in solution, it adds Cl鈦 ions, thereby increasing the concentration of Cl鈦 in the solution.

The added Cl鈦 ions shift the equilibrium of the AgCl dissolution process, reducing the solubility of AgCl. This is because more Cl鈦 ions in the solution push the equilibrium towards the left, encouraging the formation of solid AgCl. Hence, fewer Ag鈦 ions dissolve, which is why the solubility of AgCl becomes much lower than in pure water."}]} 褉邪唳邪谢 academy assistant arbeitet canThe canakinumab銇椼仸銇忋仩銇曘亜Campiones 瓿犽鞀惦媹雼.鈻 solicitam,n.Braune Ein NO璨 欤 懈屑銉兗 谐邪写卸械褌褘 riall'attive脟A Bukowski

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Most popular questions from this chapter

\(0.01\) liter of \(0.3 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) is mixed with \(0.02\) liters of a solution that is initially \(0.1 \mathrm{M} \mathrm{ca}^{++}\) and \(0.1 \mathrm{M} \mathrm{Sr}^{++}\). Given that \(\mathrm{K}_{s p}=2.4 \times 10^{-5}\) for \(\mathrm{CaSO}_{4}\) and \(\mathrm{K}_{\mathrm{sp}}=7.6 \times 10^{-7}\) for \(\mathrm{SrSO}_{4}\) what is the final concentration of \(\mathrm{Ca}^{++}, \mathrm{Sr}^{++}\), and \(\mathrm{SO}^{--} 4\) in this solution?

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A chemist has a saturated solution of \(\mathrm{CaSO}_{4}\) in \(.0100 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). The \(\mathrm{K}_{\mathrm{sp}}\) of \(\mathrm{CaSO}_{4}\) is \(2.4 \times 10^{-5} .\) calculate the concentration of \(\mathrm{Ca}^{2+}\) ion in this saturated solution.

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