For each of the following compounds (here shown in their acidic forms), write the form that predominates in a solution with a pH = 5.5:
i. HON+H3 (pKa = 6.0)
For each of the following compounds (here shown in their acidic forms), write the form that predominates in a solution with a pH = 5.5:
i. HON+H3 (pKa = 6.0)
Using pKₐ values, calculate the equilibrium constants for the following acid–base reactions.
(b)
Amino acids exist predominantly in one of the following forms. Which is it? Explain your answer.
Which of the following is a correct statement about the equilibrium constant () in a chemical reaction?
At what pH is the concentration of a compound, with a pKa = 8.4, 100 times greater in its acidic form than in its basic form? At what pH is 50% of a compound, with a pKa = 7.3, in its basic form?
For the following acid–base pairs, (iv) calculate Keq;
(e)
For each of the following compounds, indicate the pH at which
a. 50% of the compound is in a form that possesses a charge.
1. CH3CH2COOH (pKa = 4.9)
2. CH3N+H3 (pKa = 10.7)
Hydrogen gas (H2) has a relatively high pKa value. Is it a stable or unstable acid? Do you expect it to participate in acid–base reactions?
A naturally occurring amino acid such as alanine has a group that is a carboxylic acid and a group that is a protonated amine. The pKa values of the two groups are shown.
d. Draw the structure of alanine in a solution at pH = 12.
At what pH does 80% of the acid exist in its acidic form?
Write the Keq expression for the following acid–base reactions. [You don't need to calculate Keq here.]
(c)
Using pKa values for the conjugate acids of the bases on each side of the reaction arrow, identify which side of the equilibrium would be favored in the following hypothetical reactions.
(c)
We usually calculate Keq for acid–base reactions using pKa values.
(a) Derive an equation to calculate Keq using pKb values, then
(b) use it to calculate the equilibrium constant for the following reaction.
For the following acid–base reaction, (b) calculate the equilibrium constant.