Find the pH for a solution that is 0.5 M in NH4Cl and 2.0 M in NH3 when the Kb= 1.75*10-4 for NH3 . NH4Cl is a salt that ionizes to NH4+ and Cl-.
The accepted Kb value for NH3 is 1.75 x 10-5 but you have given the value 1.75 x 10-4 . So I will calculate the pH with your value
NH4Cl is an ionic salt which completely dissociates to NH4+ and Cl-
Concentration of NH4+ = concentration of NH4Cl
Concentration of NH4+ = 0.5 M
Kb = 1.75 x 10-4
pKb = -log(Kb)
pKb = -log(1.75 x 10-4)
pKb = 3.76
According to Henderson-Hasselbalch equation
pOH = pKb + log([conjugate acid] / [weak base])
where [conjugate acid] = concentration of conjugate acid = [NH4+] = 0.5 M
[weak base] = concentration of weak base = [NH3] = 2.0 M
pKb = 3.76
Substituting these values,
pOH = 3.76 + log(0.5 M / 2.0 M)
pOH = 3.76 - 0.60
pOH = 3.16
pH = 14 - pOH
pH = 14 - 3.16
pH = 10.84
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