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.

pKa = - 1.36012

Explanation:

∴ Ka = [H+] [A-] / [HA]

∴ %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

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the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.