The density of water is equal to 1.0 g/cm3. a substance with density equal to 0.9 g/cm3 would _____ in water.

The equilibrium constant for ammonia[tex]\left({{{\text{K}}_{\text{b}}}}\right)[/tex] is[tex]\boxed{1.82\times{{10}^{-5}}}[/tex].

Further explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

[tex]{\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}[/tex]

Equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for general reaction is as follows:

[tex]{\text{K}}=\frac{{\left[{\text{D}}\right]\left[{\text{C}}\right]}}{{\left[{\text{A}}\right]\left[{\text{B}}\right]}}[/tex]

Here, K is the equilibrium constant.

The equilibrium constant for the dissociation of acid is known as [tex]{{\text{K}}_{\text{a}}}[/tex]and equilibrium constant for the dissociation of base is known as[tex]{{\text{K}}_{\text{b}}}[/tex].

The expression that relates pH and pOH is given as follows:

[tex]{\text{pH}}+{\text{pOH}}=14[/tex]                                   …… (1)

Rearrange equation (1) to calculate pOH.

[tex]{\text{pOH}}=14-{\text{pH}}[/tex]                                 …… (2)

Substitute 11.50 for the value of pH in equation (2).

[tex]\begin{aligned}{\text{pOH}}&=14-{\text{11}}{\text{.50}}\\&={\text{2}}{\text{.5}}\\\end{aligned}[/tex]

pOH is the measure of hydroxide ion concentration. The formula to calculate pOH is as follows:

[tex]{\text{pOH}}=-\log\left[{{\text{O}}{{\text{H}}^-}}\right][/tex]                              …… (3)

Here,

[tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex] is the concentration of hydroxide ion.

Rearrange equation (3) to calculate [tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex].

[tex]\left[{{\text{O}}{{\text{H}}^-}}\right]={10^{-{\text{pOH}}}}[/tex]                              …… (4)

Substitute 2.5 for pOH in equation (4).

[tex]\begin{aligned}\left[{{\text{O}}{{\text{H}}^-}}\right]&={10^{-2.5}}\\&=0.0031622\\\end{aligned}[/tex]

The given equilibrium reaction is,

[tex]{\text{N}}{{\text{H}}_{\text{3}}}+{{\text{H}}_2}{\text{O}}\rightleftharpoons{\text{NH}}_4^++{\text{O}}{{\text{H}}^-}[/tex]

The expression of [tex]{{\text{K}}_{\text{b}}}[/tex]for the above reaction is as follows:

[tex]{{\text{K}}_{\text{b}}}=\frac{{\left[{{\text{NH}}_4^+}\right]\left[{{\text{O}}{{\text{H}}^-}}\right]}}{{\left[{{\text{N}}{{\text{H}}_3}}\right]}}[/tex]                                  …... (5)

The equilibrium concentration of both [tex]{\text{NH}}_4^+[/tex] and [tex]{\text{O}}{{\text{H}}^-}[/tex] is the same.

0.0031622 M of [tex]{\text{O}}{{\text{H}}^-}[/tex]is present at equilibrium so 0.0031622 M out of 0.55 M of  has reacted.

The initial concentration of the aqueous solution is 0.55 M. So the concentration of [tex]{\text{N}}{{\text{H}}_3}[/tex] left at equilibrium is calculated as follows:

[tex]\begin{aligned}\left[{{\text{N}}{{\text{H}}_3}}\right]&={\text{Initial concentration of N}}{{\text{H}}_{\text{3}}}-{\text{Reacted concentration of N}}{{\text{H}}_{\text{3}}}\\&={\text{0}}{\text{.55 M}}-{\text{0}}{\text{.0031622 M}}\\&={\text{0}}{\text{.5468378 M}}\\\end{aligned}[/tex]

The value of[tex]\left[{{\text{O}}{{\text{H}}^-}}\right][/tex]is 0.0031622 M.

The value of [tex]\left[{{\text{N}}{{\text{H}}_3}}\right][/tex] is 0.5468378 M.

The value of [tex]\left[{{\text{NH}}_4^+}\right][/tex] is 0.0031622 M.

Substitute these values in equation (5).

[tex]\begin{aligned}{{\text{K}}_{\text{b}}}&=\frac{{\left({0.0031622\;{\text{M}}}\right)\left({0.0031622\;{\text{M}}}\right)}}{{\left({0.546837{\text{8 M}}}\right)}}\\&=1.82861\times{10^{-5}}\\&\approx1.82\times{10^{-5}}\\\end{aligned}[/tex]

Therefore, equilibrium constant for ammonia is[tex]{\mathbf{1}}{\mathbf{.82\times1}}{{\mathbf{0}}^{{\mathbf{-5}}}}[/tex].

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: https://brainly.com/question/3467841

2. Complete equation for the dissociation of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex](aq): https://brainly.com/question/5425813

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Equilibrium

Keywords: NH3, OH-, NH4+, H2O, equilibrium, kb, pH, pOH, 14, 11.5, 2.5, aqueous solution, 0.0031622 M, 0.5468378 M.

The Molarity concentration is expressed in units of moles / L. So let us first determine the number of moles of CH3OH, and then divide that amount by the total volume of 0.230 L of solution.

To determine the number of moles of CH3OH, divide the weight in grams of CH3OH by the molecular weight of CH3OH: (MW of CH3OH = 32 g / mol)

number of moles = 16.8 g / (32 g / mol)

number of moles = 0.525 mol CH3OH 

Then we calculate for molarity:

Molarity = 0.525 mol CH3OH / .230 L

Molarity = 2.2826 mol / L

Molarity = 2.28 M

The molarity of a solution prepared by dissolving 16.8 g of CH₃OH in sufficient water to give exactly 230 mL of solution is 2.28 M.

The concentration of CH₃OH in a solution can be determined by calculating its molarity, which is the number of moles of solute per liter of solution. To find the molarity, we first need to convert the mass of CH₃OH to moles. The molar mass of CH₃OH (Methanol) is 32.04 g/mol.

Step 1: Convert the mass of CH₃OH to moles using its molar mass.
16.8 g CH₃OH ×1 mol CH₃OH / 32.04 g CH₃OH = 0.524 moles CH₃OH

Step 2: Convert the volume of the solution from milliliters to liters.
230 mL ×1000 mL/L = 0.230 L

Step 3: Calculate the molarity of the solution.
0.524 moles CH₃OH / 0.230 L = 2.28 M