Question

A 0.21 M solution of a weak acid HA dissociates such that 99.1% of the weak acid remains intact (i.e., remains as HA). To the nearest hundredths, what is the pKa of this weak acid? Assume the temperature is 25 °C. Also, the value of R, the gas constant, is 1.987 cal/KLaTeX: \cdot⋅mol.

Answer 1

Answer:

pKa = - 1.36012

Explanation:

• HA ↔ H+ + A-

∴ Ka = [H+] [A-] / [HA]

• pKa = - Log Ka

• % dissociates (%α) = ([A-] / [A-] + [HA])×100

∴ %α = 99.1%

∴ C HA = 0.21 M

Mass Balance:

⇒ C HA = [HA] + [A-] = 0.21 M............(1)

Charge balance:

⇒ [H+] = [A-] + [OH-].....[OH-] is neglected, it come from the water

⇒ [H+] = [A-]...............(2)

(2) in (1):

⇒ [HA] = 0.21 - [H+]

replacing in %α:

∴ %α = 99.1% = ([A-]/([A-]+[HA]))×100

⇒ 0.991 = [A-] / [A-] + [HA] = [A-] / 0.21 M

⇒ [A-] = (0.21 M)*(0.991) = 0.20811 M

Replacing in Ka:

⇒ Ka = [H+]² / (0.21 - [H+])

∴ [H+] = [A-] = 0.20811 M

⇒ Ka = (0.20811)² / (0.21 - 0.20811)

⇒ Ka = 22.9152

⇒ pKa = - Log (22.9152)

⇒ pKa = - 1.3601