Bromine, a liquid at room temperature, has a boiling point of 58C and a melting point of -7.2C. Bromine can be classified as a
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HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:
HClO + H2O ⇄ H3O+ +OCl-
Then the equilibrium constant, Ka, of dilute HClO would be:
![K_{a} = \frac{[ H_{3} O^{+} ][O Cl^{-} ]}{HClO}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5B%20H_%7B3%7D%20%20O%5E%7B%2B%7D%20%5D%5BO%20Cl%5E%7B-%7D%20%5D%7D%7BHClO%7D%20)
Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x
Substituting the excess (E) concentration to the Ka equation:
![K_{a} = \frac{[x ][x]}{1.0 \ x \ 10^{-4} - x }](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5Bx%20%5D%5Bx%5D%7D%7B1.0%20%5C%20x%20%5C%20%2010%5E%7B-4%7D%20-%20x%20%7D)
Simplifying the equation would yield a quadratic equation:
The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.
x = 1.688 x 10^-6 M = [H3O+] = [ClO-]
Then, you can determine the conc of [OH-] through pH.
pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M
Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M
Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.
Thus in relative amounts, the order would be
H2O >>> HClO > H3O+ = ClO- > OH-
HClO + H2O ⇄ H3O+ +OCl-
Then the equilibrium constant, Ka, of dilute HClO would be:
Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x
Substituting the excess (E) concentration to the Ka equation:
Simplifying the equation would yield a quadratic equation:
The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.
x = 1.688 x 10^-6 M = [H3O+] = [ClO-]
Then, you can determine the conc of [OH-] through pH.
pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M
Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M
Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.
Thus in relative amounts, the order would be
H2O >>> HClO > H3O+ = ClO- > OH-
Answer:
See explanation and image attached
Explanation:
This reaction is known as mercuric ion catalyzed hydration of alkynes.
The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.
The IUPAC rules are
a) Find out the longest chain of carbon in the given organic compound
b) We will name the longest chain.
c) We will identify the main functional group and will assign a suffix to the compound.
d) We will number the carbons in the longest chain selected so that the attached groups attain lowest numeral as substituent
e) We will name the side groups or chains.
