determine the pH of 0.15M ammonia assuming the dissociation value of ammonia is Kb=1.8*10^-5

To determine the pH of a solution of ammonia, we need to first understand its chemical properties and how it dissociates in water.

Ammonia, NH3, is a weak base. When dissolved in water, it undergoes partial dissociation, producing hydroxide ions (OH-) and ammonium ions (NH4+).

The equilibrium reaction for the dissociation of ammonia can be represented as follows:

NH3 + H2O ⇌ NH4+ + OH-

The dissociation constant, Kb, is a measure of the extent to which the reaction proceeds. For ammonia, the value of Kb is given as 1.8 × 10^-5.

The relationship between Kb and Kw (the ion product of water) can be used to find the concentration of hydroxide ions (OH-) produced by the dissociation of ammonia.

Kb = [NH4+][OH-]/[NH3]

Since we know the concentration of ammonia (0.15 M) and the value of Kb, we can rearrange the equation and solve for [OH-].

[OH-] = Kb * [NH3] / [NH4+]

Substituting the values into the equation:

[OH-] = (1.8 × 10^-5) * (0.15) / (0.15)

Simplifying further:

[OH-] = 1.8 × 10^-5

Now, to calculate the pH, we need the concentration of hydroxide ions ([OH-]). From the above calculation, we found [OH-] to be 1.8 × 10^-5 M.

pOH is calculated by taking the negative logarithm (base 10) of the hydroxide ion concentration:

pOH = -log[OH-]
= -log(1.8 × 10^-5)
= 4.74

Finally, we can use the pH formula to find the pH:

pH = 14 - pOH
= 14 - 4.74
= 9.26

Therefore, the pH of a 0.15 M ammonia solution is approximately 9.26.