Question

1a) What is the pH of a 0.225 mol L−1 ammonia solution?

b) What is the percent ionization of ammonia at this concentration?

2) A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries.

To find the pKa of X-281, you prepare a 0.095 mol L−1 test solution of X-281. The pH of the solution is determined to be 3.00. What is the pKa of X-281?

Answer 1

NH3 + H2O → NH4+ + OH-
let [OH-] = X then [NH4+] = X and {NH3] = 0.225-X

Kb = 1.8E-5
Kb = [NH4+] [OH-] / [NH3]
1.8E-5 = X^2 / (0.225-X)
4.05E-6 - 1.8E-5 X = X^2
X^2 + 1.8E-5 X - 4.05E-6 = 0
X = 2.00E=3
p(OH) = -log(2.00E-3) = 2.70
pH = 14-2.70= 11.3

b
X/100 = ionized / original =2.00E-3 / 0.225
X = 0.89%

1b) 15.8% ionised

2) When a weak acid is dissolved in water it reaches an equilibrium:

HA = H^+ + A^-, where for each HA ionized one H^+ and one A^- are produced.

[HA] = 2.52 x 10^-2 M, and [H^+] = [A^-]
pH = 3.0 = -log[H^+]
[H^+] = antilog 10^-3.0 = 1.175 x 10^-3 M

Use the Henderson-Hasselbalch equation: pKa = pH - log[A-]/[HA] {Notice, I rearranged this equation to fit my need to obtain pKa.}
pKa = 3.0 - log(1.175 x 10^-3)/(2.52 x 10^-2) = 3.0 - (-1.33) = 4.33.