Liquid NH₃ ionises to a slight extent. At a certain temperature it's self ionization constant K_SIC(NH3) = 10^-30. The number of NH₄⁺ ions are present per 100 cm³ of pure liquid are??

To determine the number of NH4+ ions present in 100 cm³ of pure liquid ammonia (NH3) at a given temperature, we can use the self-ionization constant (K SIC) of ammonia. The self-ionization of ammonia can be represented by the following equilibrium reaction:

2 NH3 ⇌ NH4+ + NH2-

Given that the self-ionization constant K SIC(NH3) is 10^-30, we can set up an expression for K SIC based on the concentrations of the ions produced at equilibrium.

Understanding the Ionization Process

In this reaction, for every two molecules of ammonia that ionize, one ammonium ion (NH4+) and one amide ion (NH2-) are produced. If we let 'x' represent the concentration of NH4+ ions at equilibrium, then the concentration of NH2- ions will also be 'x'. The concentration of un-ionized NH3 will be reduced by 2x, since two molecules of NH3 are consumed for every ion pair produced.

Setting Up the Equilibrium Expression

The equilibrium expression for K SIC can be written as:

K SIC = [NH4+][NH2-] / [NH3]^2

Substituting the concentrations in terms of 'x', we have:

K SIC = (x)(x) / (C - 2x)^2

Where C is the initial concentration of NH3. However, since NH3 is a pure liquid, we can approximate its concentration as a constant. For pure liquid ammonia, the density is approximately 0.682 g/cm³, and the molar mass of NH3 is about 17 g/mol. Therefore, the concentration of NH3 can be calculated as follows:

• Density of NH3 = 0.682 g/cm³
• Molar mass of NH3 = 17 g/mol
• Concentration (C) = Density / Molar mass = 0.682 g/cm³ / 17 g/mol ≈ 0.0401 mol/L

Calculating the Ion Concentration

Now, substituting the values into the K SIC expression:

10^-30 = (x)(x) / (0.0401 - 2x)^2

Assuming that 2x is negligible compared to 0.0401 (which is valid since K SIC is very small), we can simplify the equation to:

10^-30 = x² / (0.0401)²

Rearranging gives us:

x² = 10^-30 * (0.0401)²

x² = 10^-30 * 1.60801 x 10^-3 ≈ 1.60801 x 10^-33

Taking the square root of both sides:

x ≈ 1.27 x 10^-17 mol/L

Finding the Number of Ions in 100 cm³

Now, to find the number of NH4+ ions in 100 cm³ (which is 0.1 L), we can use the concentration we just calculated:

Number of moles of NH4+ = concentration × volume = (1.27 x 10^-17 mol/L) × (0.1 L) = 1.27 x 10^-18 mol

To convert moles to the number of ions, we use Avogadro's number (approximately 6.022 x 10²³ ions/mol):

Number of NH4+ ions = (1.27 x 10^-18 mol) × (6.022 x 10²³ ions/mol) ≈ 7.65 x 10^5 ions

Final Result

Therefore, in 100 cm³ of pure liquid ammonia, there are approximately 765,000 NH4+ ions present. This illustrates how even in a pure liquid, the extent of ionization can lead to a significant number of ions, albeit still a very small concentration overall.