Anew 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 0.089 m test solution of x-281 at 25.0 ∘c. the ph of the solution is determined to be 2.40. what is the pka of x-281?

The pKa of X-281 is 3.73.

Explanation:

X-281 is a monoprotic weak acid so that it will not dissociate completely in solution (weak acid) and will only produce 1 mol of protons per mol compound (monoprotic acid).

The dissociation equation could be written as follows:

X-281-H ⇆ X-281 + H⁺

Note the equilibrium arrows indicating that not all X-281-H dissociates at equilibrium.

Initially, the concentration of X-281-H is 0.089 M. At equilibrium, the concentration of the dissociation products, X-281 and H⁺, is unknown but both must be the same, since the drug is a monoprotic acid. We can call "X" to the concentration of the products. The concentration of X-281-H is the initial concentration minus the concentration of the products: 0.089 M - X. Then, at equilibrium, these are the concentrations of each species present:

  X-281-H               ⇆      X-281        +       H⁺

(0.089 M-X)                         X                    X

We also know that the pH is 2.40. Then:

pH = -log[H⁺] = 2.40

where [H⁺] is the molar concentration of the protons or "X". Then:

-log X = 2.40

log X = -2.40

X = 10^(-2.40) = 3.98 x 10⁻³ M

The concentration of each species present in the equilibrium is then:

[H⁺] = 3.98 x 10⁻³ M

[X-281] = 3.98 x 10⁻³ M

[X-281-H] = 0.089 M - 3.98 x 10⁻³M = 0.085 M

At equlibrium, the acidity constant Ka is:

Ka = [X-281] * [H⁺] / [X-281-H]

Ka = (3.98 x 10⁻³ M * 3.98 x 10⁻³ M ) / 0.085 M = 1.86 x 10⁻⁴

Then the pKa is:

pKa = -log Ka = -log (1.86 x 10⁻⁴) = 3.73

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%