a. Calculate the molar solubility of AgBr in pure water. Ksp for AgBr is 5.0 ×10^-13 b. Calcula

A. Calculate the molar solubility of AgBr in pure water. Ksp...

a. Calculate the molar solubility of AgBr in pure water. $K_{\text {sp }}$ for AgBr is $5.0 \times 10^{-13}$ b. Calculate the molar solubility of AgBr in $3.0 M \mathrm{NH}_{3}$. The overall formation constant for $\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}$ is $1.7 \times 10^{7}$, that is, $\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of $\mathrm{AgBr}$ will dissolve in $250.0 \mathrm{~mL}$ of $3.0 \mathrm{M} \mathrm{NH}_{3}$ ? e. What effect does adding $\mathrm{HNO}_{3}$ have on the solubilities calculated in parts a and $\mathrm{b}$ ?

a. Calculate the molar solubility of AgBr in pure water. $K_{\text {sp }}$ for AgBr is $5.0 \times 10^{-13}$
b. Calculate the molar solubility of AgBr in $3.0 M \mathrm{NH}_{3}$. The overall formation constant for $\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}$ is $1.7 \times 10^{7}$, that is,
$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$
c. Compare the calculated solubilities from parts a and b. Explain any differences.
d. What mass of $\mathrm{AgBr}$ will dissolve in $250.0 \mathrm{~mL}$ of $3.0 \mathrm{M} \mathrm{NH}_{3}$ ?
e. What effect does adding $\mathrm{HNO}_{3}$ have on the solubilities calculated in parts a and $\mathrm{b}$ ?

Key Concepts

Molar Solubility

Molar solubility is defined as the number of moles of a solute that can dissolve in one liter of solution to form a saturated solution. It is determined by applying the solubility product expression and considering any equilibrium shifts due to additional factors such as complex ion formation or the presence of common ions.

Quantitative Conversion for Mass Calculations

Quantitative conversion involves using the molar solubility to determine the number of moles of a substance dissolved in a given volume of solution and subsequently converting this amount to mass using the compound's molar mass. This approach is essential for practical calculations where the final goal is to understand the actual quantity of solute that can be dissolved under the specified conditions.

Le Chatelier's Principle and Equilibrium Shifts

Le Chatelier's Principle explains how a chemical equilibrium responds to disturbances such as changes in concentration, pH, or the addition of complexing agents. In the context of solubility equilibria, this principle helps predict how the addition of ligands, acids, or bases will affect the balance of production and dissolution of ions, thereby influencing the molar solubility of the salt.

Complex Ion Formation (Formation Constant, Kf)

Complex ion formation occurs when a metal ion binds with one or more ligands to form a complex ion, which can significantly alter the solubility of a salt. The formation constant, or Kf, quantifies the stability of the complex ion, and a high Kf value indicates a strong tendency for complexation. This process reduces the concentration of free metal ions in solution, thereby shifting the equilibrium and increasing the overall solubility of the sparingly soluble salt.

Solubility Product (Ksp)

The solubility product is an equilibrium constant specific to the dissolution of a sparingly soluble salt. It represents the product of the molar concentrations of the constituent ions each raised to the power of their stoichiometric coefficients in a saturated solution. This constant is used to calculate the molar solubility and to predict whether a precipitate will form under given conditions when the ion product exceeds the Ksp value.

Transcript

00:02 Starting with option a, the system initially contains water and solid silver bromide which dissolve in water to give silver iron and bromide negative iron.

00:17 A .g.

00:19 Br gives a .g.

00:21 Positive plus b .r.

00:25 The solubility product constant, ksp, will be equal to concentration of products, silver and bromine.

00:35 Negative ion.

00:37 To find the solubility we need to list out initial and equilibrium concentration.

00:48 So starting with ag positive 0 and bromine negative it is also 0.

00:59 And for the equilibrium one it will be x and x move per liter.

01:09 Now we substitute the equilibrium concentration into equilibrium constant expression.

01:16 That goes ksp equals to 5 .0 multiplied 10 to 1 negative 13 equals x multiplied with x.

01:29 So value of x turn out to be 7 .1 multiplied 10 3 % 0 .m .1 multiplied 10 raised to 1 negative 7 .m.

01:37 Option b.

01:41 Let me start with the writing the reaction.

01:50 Ag br gives ag positive plus br negative this is the first reaction they go to 7 .7 multiply 10 raised to power negative 13 the second reaction is ag positive plus 2nh3 gives me ag nh32 positive this is the second reaction if s is taken as mole per liter of ag be dissolved then s will be equal to the concentration of bromine negative ion which is almost equivalent to concentration of ag nh3 to positive iron and h3 concentration will be equal to 1 .0 minus 2 s this is because 2 moles of nh3 is used for for forming a complex.

03:08 S squared divided by 3 .00 minus 2 s whole square...

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