Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
![E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}](https://tex.z-dn.net/?f=E%20%3D%20E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D)
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
![E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20%20%3D%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%5C%5C%5C%5Clog%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%3D%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%20%5C%5C%5C%5Clog%5Bred%5D%20%3D%20%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%7D%20%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log0.733%20-%20%20%5Cfrac%7B2x5.45x10%5E%7B-2%7D%20%20%7D%7B0.0591%7D%7D%5C%5C%5C%5C)
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.
There are 10 hydrogen atoms that bind and there are 2 pairs of free electrons in the non-binding O atom
<h3>Further explanation</h3>
Aldehydes are alkane-derived compounds containing carbonyl groups (-CO-) where one bond binds to an alkyl group while another binds to a hydrogen atom.
The general structure is R-CHO with the molecular formula :
Naming is generally the same as the alkane by replacing the suffix with -al
Butanal or butyraldehyde is an aldehyde which has 4 C atoms
Inside the structure there are 3 atoms involved in bonding:
- 1. Atom C with 4 valence electrons, requires 4 electrons to reach the octet
- 2. Atom O with 6 valence electrons, requires 2 electrons to reach the octet
- 3. Atom H with 1 valence electron, requires 1 electron to reach a duplet
In describing Lewis's structure the steps that can be taken are:
- 1. Count the number of valence electrons from atoms in a molecule
- 2. Give each bond a pair of electrons
- 3. The remaining electrons are given to the atomic terminal so that an octet is reached
- 4. The remaining electrons that still exist in the central atom
- 5. If the central atom is not yet octet, free electrons are drawn to the central atom to form double bonds
In the Butanal structure (C₄H₈O) there is 1 double bond of the functional group (-CHO) between the C atom and the O atom
<h3>Learn more:
</h3>
Adding electron dots
brainly.com/question/6085185
Ionic bonding
brainly.com/question/1603987
Formal charge
brainly.com/question/7190235
Keywords: butanal, aldehyde, Lewis structure, a valence electron
It's a mixture of H2O + a soluble substance like, salt.
The molarity of the solution is 0.01.
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