Chemistry, 27.09.2019 15:00 scastillo8

The ph of a 0.55 m aqueous solution ammonia, nh3, at 25.0°c is 11.50. what is the value of kb for nh3?

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Answer from: mdkester11

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

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:

${\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}$

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:

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

Here, K is the equilibrium constant.

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

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

${\text{pH}}+{\text{pOH}}=14$ …… (1)

Rearrange equation (1) to calculate pOH.

${\text{pOH}}=14-{\text{pH}}$ …… (2)

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

$\begin{aligned}{\text{pOH}}&=14-{\text{11}}{\text{.50}}\\&={\text{2}}{\text{.5}}\\\end{aligned}$

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

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

Here,

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

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

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

Substitute 2.5 for pOH in equation (4).

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

The given equilibrium reaction is,

${\text{N}}{{\text{H}}_{\text{3}}}+{{\text{H}}_2}{\text{O}}\rightleftharpoons{\text{NH}}_4^++{\text{O}}{{\text{H}}^-}$

The expression of ${{\text{K}}_{\text{b}}}$ for the above reaction is as follows:

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

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

0.0031622 M of ${\text{O}}{{\text{H}}^-}$ 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 ${\text{N}}{{\text{H}}_3}$ left at equilibrium is calculated as follows:

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

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

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

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

Substitute these values in equation (5).

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

Therefore, equilibrium constant for ammonia is ${\mathbf{1}}{\mathbf{.82\times1}}{{\mathbf{0}}^{{\mathbf{-5}}}}$ .

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c:

2. Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ (aq):

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.

Answer from: Quest

but know think of it in terms of bond breaking and bond making.

h–h + h–h + o=o ==> h–o–h + h–o–h

as you can see from the diagram, theoretically, more energy is released by forming water, than was absorbed in breaking up the original hydrogen and oxygen molecules.

the energy level of the fuel + oxygen is higher than that of water.

therefore the fuel + oxygen mixture has the potential to release potential chemical energy as heat energy or, crucially, electrical energy, if the reaction can be made to go via oxidation and reduction reactions involving electron transfer.

see this page for more details of the full bond energy calculations

a fuel cell is a device to make electricity that functions like an electrical cell or battery that you supply with a fuel (liquid or gas) and an oxidant (oxidising agent, usually oxygen gas) which react together in a redox reaction (oxidation + reduction) to release energy as a flow of electrons i.e. an electrical current capable of doing useful work.

the oxidation of the fuel and the reduction of the oxidant reactions take place on electrodes (see diagram on right, all fully explained further down the page).

fuel cells essentially a device to release the chemical potential energy of combustible fuels as electrical energy at a much lower temperature and without the flame!

so, the idea of a hydrogen–oxygen fuel cell is to release the energy from hydrogen reacting with oxygen, not as heat as in normal combustion in air, but as useful electrical energy i.e. a practical electricity supply.

hydrogen-oxygen fuel cells are much efficient than conventional power stations or batteries (e.g. zinc-carbon) because the electrical energy is directly generated from the chemical reaction between the oxidant and the fuel - there are no complications with boilers, turbines and generators.

in fossil fuel power stations, motor vehicles, coal/gas fires etc. quite a percentage of energy is lost as waste heat.

with a fuel cell there are fewer stages in producing the useful energy, so there is less opportunity to lose potentially useful energy e.g. waste heat, friction from moving parts etc.

a hydrogen–oxygen fuel cell is a non–polluting clean fuel since the only combustion product is water.

fuel cells do not produce pollutants like carbon monoxide, sulfur dioxide and nitrogen oxides - gases you get from fossil fuel combustion in power stations or burnt road vehicles..

cars powered by fuel cells would be quite an environmental advantage in cities, where electric cars are already beginning to be significantly used in developed countries.

fuel cells could replace larger batteries which are not easily recycled and contain highly toxic metal compounds.

it would be an ideal fuel on this basis e.g. for motor vehicles, but that's not the only factor to consider!

it would be ideal if the hydrogen fuel could be manufactured by electrolysis of water e.g. using solar cells.

hydrogen can be used to power fuel cells.

it all sounds wonderful but, still technological problems to solve for large scale manufacture and distribution of 'clean' hydrogen gas or use in generating electricity and its rather an inflammable explosive gas!

explanation:

fuel cells e.g. the hydrogen-oxygen fuel electrical cell

fuel cells have to be supplied by an external source of fuel (e.g. hydrogen) and an oxidant e.g. oxygen or air.

the hydrogen or any other fuel is oxidised electrochemically inside the fuel cell to produce a potential difference i.e. a voltage capable of producing a working current.

the overall chemical reaction in a hydrogen fuel electrochemical cell involves the oxidation of hydrogen by oxygen to produce only water.

hydrogen fuel cells offer an alternative to rechargeable cells and batteries.

a fuel cell will produce a potential difference ('voltage') and a workable electric current until one of the reactants is used up.

hydrogen gas can be used as fuel.

it burns with a pale blue flame in air reacting with oxygen to be oxidised to form water.

hydrogen + oxygen ==> water

2h2(g) + o2(g) ==> 2h2o(l)

Answer from: Quest

what is the average time it took the car to travel 0.25 meters?

⇒ 2.23 seconds

what is the average time it took the car to travel 0.50 meters?

⇒ 3.13 seconds

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