A) pH of Buffer Mix:
(HO)3-CNH4OH <=> (HO)3-CNH4Cl + OH
[(HO)3-CNH4OH] = [(10g/97g/mol)/(0.250L)=0.412M
[(HO)3-CNH4OHCl] = [(10g/115g/mol)/(0.250L)=0.348M
Kb(fm table of wk base ionization constants) = 1.202x10^-6
=> Ka = Kw/Kb = (1.0x10^-14/1.202x10^-6) = 8.32x10^-9
=> pKa = -log(Ka) = -log(8.32x10^-9) = 8.08
Using Henderson–Hasselbalch equation:
pH(Buffer) = pKa + log([Base]/[Acid])
pH(Buffer) = 8.08 + log[(0.412)/(0.348)] = 8.08 + 0.07 = 8.15
B) pH after adding 10.5ml(0.500M NaOH)
=> moles NaOH added = (0.500M)(0.0105L) = 0.0053 mole NaOH
[NaOH]added = [OH]added = [(0.0053mole)/(0.250+0.0105)L] = [(0.0053)/(0.2605)]M = 0.0202M(OH^-)excess
New Equilibrium ... Excess OH^- is removed by Cation + OH^- => Cat-OH; rxn shifts toward the 'Base' side of equilibrium.
Left Shift increases [Base] => Add excess [OH] to reactant side, and decreases [Acid] => Subtract [OH] from product side...
=>(HO)3-CNH4OH<=>(HO)3-CNH4^+ + OH^-
Ci-- 0.412M -------- 0.348M ----(~0)
∆C--+0.020M ------- -0.020M --------
Ceq-0.4322M ------- 0.328M ----(x)
Kb=[(HO)3-CNH4^+][OH^-]/[(HO)3-CNH4OH]
=> [OH](new) = [Kb[(HO)3-CNH4OH)]/[(HO)3-CNH4^+]
=> [OH](new) = [(1.202x10^-6)(0.4322)/(0.3278)] = 1.58x10^-6
=> pOH = -log[OH] =-log(1.58x10^-6) = 5.8
=> pH(new) = 14 - pOH = 14 - 5.8 = 8.20 (pH increases slightly as expected with base addition).
Calculate the pH of a solution prepared by dissolving 10.0 g of tris(hydroxymethyl)aminomethane ("tris base") plus 10.0 g of tris hydrochloride in 0.250 L of water.
(a) What will be the pH if 10.5 mL of 0.500 M NaOH is added?
3 answers
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