When total ammonium ion is 50% dissociated, The pH is 9.25.
Given, the pKₐ of ammonium ion (NH₄⁺) is 9.25.
The dissociation of NH₄⁺ to ammonia (NH₃) and H⁺ in aqueous solution is represented below.
It is stated that the ammonium ion is 50% dissociated in the solution. Therefore the concentration of each species at that instant [if started with 1M conc. of NH4+ (aq)] is summarized as follows:
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
Initial conc. (in M) 1 - -
At 50% dissociation 1-0.5, i.e., 0.5 0.5 0.5
The given value of pKₐ immediately reminds us of the form of Henderson-Hasselbalch equation containing pKₐ , i.e.,
pH = pKₐ + log₁₀ [tex]\frac{[salt]}{[acid]}[/tex] _ _ _(i)
But, NH₄⁺ (aq) exists as the conjugate acid form, as well as the salt (ionized) form. Hence, we cannot use equation. (i) for our calculations.
Therefore, we need to convert the above value of Kₐ to the equivalent Kb (for NH₃), corresponding to the reverse equilibrium.
This is achieved by using the relation
Kb = [tex]\frac{Kw}{Ka}[/tex]
or, -log₁₀ Kb = -log₁₀ Kw + log₁₀ Ka
or, pKb = pKw - pKa = 14 - 9.25 = 4.75
where, pKw = 1 x 10⁻¹⁴ (at 298 K).
Therefore, pKb of ammonia (NH₃) = 4.75
Now, we need to use the form of Henderson-Hasselbalch equation which uses Kb; which is in fact
pOH = pKb + log₁₀ [tex]\frac{[salt]}{[base]}[/tex] _ _ _(ii)
where the salt is NH₄⁺, and the base is NH₃. Further, we know that
Kw = [H⁺][OH⁻] or, pOH + pH = pKw = 14 or, pOH = 14 - pH
Substituting this relation in eq. (ii) gives
14 - pH = pKb + log₁₀ [tex]\frac{[NH4+]}{[NH3]}[/tex]
or, pH = 14 - pKb — log₁₀ [tex]\frac{[NH4+]}{[NH3]}[/tex]
As calculated, pKb = 4.75, [NH₄⁺] = 0.50 M, while [NH₃] = 0.50 M, assuming we start with a conc. of 1M NH₄⁺. Hence,
pH = 14 – 4.75 – log₁₀ [tex]\frac{0.50}{0.50}[/tex]
= 9.25 - log₁₀ ( 1 )
= 9.25 - 0
pH = 9.25
Hence, the pH of a solution consisting of 50% NH₃ and 50% NH₄⁺ will be equal to 9.25.
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