The nucleoside adenosine exists in a protonated form with a pKa of 3.8. The percentage of the protonated form at pH 4.8 is close
1 answer:
Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log() where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log() and so 10 =
or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
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Answer: The value of the equilibrium constant Kc for this reaction is 3.72
Explanation:
Equilibrium concentration of =
Equilibrium concentration of =
Equilibrium concentration of =
Equilibrium concentration of =
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
For the given chemical reaction:
The expression for is written as:
Thus the value of the equilibrium constant Kc for this reaction is 3.72