Question

Be sure to answer all parts. Calculate the following values. Enter your answers in scientific notation.

(a) What is the Kb of the benzoate ion, C6H5COO^-?
(b) What is the Ka of the 2-hydroxyethylammonium ion, HOCH2CH2NH3^+? (pKb of HOCH2CH2NH2 = 4.49)

Answer 1

To calculate the Kb of the benzoate ion, use the equation Ka x Kb = Kw and substitute the Ka of benzoic acid. The Kb of the benzoate ion is 2.38 x 10^-15. To calculate the Ka of the 2-hydroxyethylammonium ion, use the equation Ka x Kb = Kw and substitute the pKb of HOCH2CH2NH2. The Ka of the 2-hydroxyethylammonium ion is 3.98 x 10^-11.

Step-by-step explanation:

(a) To calculate the Kb of the benzoate ion, C6H5COO^-, we can use the relationship:

Ka x Kb = Kw

Since the Ka of benzoic acid is 4.20, we can substitute it into the equation to solve for Kb:

Ka x Kb = Kw

4.20 x Kb = 1.0 x 10^-14

Kb = (1.0 x 10^-14) / 4.20

Kb = 2.38 x 10^-15

(b) To calculate the Ka of the 2-hydroxyethylammonium ion, HOCH2CH2NH3^+, we can use the relationship:

Ka x Kb = Kw

Since the pKb of HOCH2CH2NH2 is given as 4.49, we can convert it to Kb and substitute it into the equation to solve for Ka:

Ka x 10^(-pKb) = Kw

Ka = Kw / 10^(-pKb)

Ka = (1.0 x 10^-14) / (10^(-4.49))

Ka = 3.98 x 10^-11