A 0.5 mol sample of he(g) of ne(g) are placed separately in two 10.0 l rigid containers at 25°c. each container has a pinhole op
1 answer:
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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
Answer:
There are present 5,5668 moles of water per mole of CuSO₄.
Explanation:
The mass of CuSO₄ anhydrous is:
23,403g - 22,652g = 0,751g.
mass of crucible+lid+CuSO₄ - mass of crucible+lid
As molar mass of CuSO₄ is 159,609g/mol. The moles are:
0,751g × = 4,7052x10⁻³ moles CuSO₄
Now, the mass of water present in the initial sample is:
23,875g - 0,751g - 22,652g = 0,472g.
mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid
As molar mass of H₂O is 18,02g/mol. The moles are:
0,472g × = 2,6193x10⁻² moles H₂O
The ratio of moles H₂O:CuSO₄ is:
2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668
That means that you have <em>5,5668 moles of water per mole of CuSO₄.</em>
I hope it helps!