A compound composed of 3.3% H, 19.3% C, and 77.4% O3.3% H, 19.3% C, and 77.4% O has a molar mass of approximately 60 g/mol.. Wha
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Answer:
2,4,4-trimethyl-2-pentene yields mixture of and
Explanation:
In ozonolysis (hydrolysis step involve a reducing agent such as Zn, etc.), a pi bond is broken to form ketone/aldehyde.
Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.
Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.
Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.
Here, 2,4,4-trimethyl-2-pentene yields mixture of and
Reaction steps are shown below.
Answer:- 0.138 M
Solution:- The buffer pH is calculated using Handerson equation:
acts as a weak acid and
as a base which is pretty conjugate base of the weak acid we have.
The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:
Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.
So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.
Let's plug in the values in the equation:
Taking antilog:
On cross multiply:
[base] = 1.38(0.10)
[base] = 0.138
So, the concentration of the base that is required to make the buffer is 0.138M.
The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium
<h3><em>Further explanation </em></h3>In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)
Acids and bases according to Bronsted-Lowry
Acid = donor (donor) proton (H + ion)
Base = proton (receiver) acceptor (H + ion)
If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.
From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.
The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant
Can be formulated:
K acid-base reaction = Ka acid on the left : K acid on the right.
or:
pK = acid pKa on the left - pKa acid on the right
K = equilibrium constant for acid-base reactions
pK = -log K;
K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.
There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:
pyridinium pKa = 5.25
acetone pKa = 19.3
butan-2-one pKa = 19
Let's look at the K value of each possible reaction:
pka H₂O = 15.74, pka of H₂CO₃ = 6.37)
- 1. C₅H₆N pyridinium
* with OH⁻
C₅H₆N + OH- ---> C₅H₅N- + H₂O
pK = pKa pyridinium - pKa H₂O
pK = 5.25 - 15.74
pK = -10.49
K values> 1 indicate the reaction can take place
* with HCO3⁻
C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃
pK = 5.25 - 6.37
pK = -1.12
Reaction can take place
- 2. Acetone C₃H₆O
* with OH-
C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O
pK = 19.3 - 15.74
pK = 3.56
Reaction does not happen
* with HCO₃-
C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃
pK = 19.3 - 6.37
pK = 12.93
Reaction does not happen
- 3. butan-2-one C₄H₇O
* with OH-
C₄H₇O + OH- ---> C₄H₆O- + H₂O
pK = 19 - 15.74
pK = 3.26
Reaction does not happen
* with HCO₃⁻
C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃
pK = 19 - 6.37
pK = 12.63
Reaction does not happen
So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium
<h3><em>Learn more </em></h3>
the lowest ph
the concentrations at equilibrium.
the ph of a solution
Keywords : acid base reaction, the equilibrium constant
