The degree of dissociation of 0.05M NH₃ at 25^oC in a solution of P^H 11 is  $p \times {10}^{-q}$
            Find the value of (p+q)?
 

Degree of Dissociation of NH3

Degree of Dissociation of 0.05M NH₃ at 25°C with pH 11

• Concentration, C = 0.05 M
• pH = 11
• pOH = 14 - 11 = 3
• [OH^-] = 10^-3 M

Degree of dissociation (α) is calculated as:

α = [OH^-] / C = 10^-3 / 0.05 = 2 × 10^-2

The value is in the form p × 10^-q where:

• p = 2
• q = 2

Therefore, the value of (p + q) = 2 + 2 = 4