Question
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?
(a) What will be the pH if 10.5 mL of 0.500 M NaOH is added?
Answers
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).
(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).
Thank you so much! you save my life
Awh Shucks! It wern't nut'n!
Related Questions
Calculate the amount of Tris(hydroxlmethyl)aminomethane(Tris) necessary to make 100mL of a 0.100 M T...
How do you prepare 100 mL of a 0.1 M TRIS-HCl buffer with a pH 7.4 using solid TRIS base (MW 121.1)...
Tris is an abbreviation for tris (hydroxymethyl) aminomethane, which is a weak base. Calculate how m...
Prepare 1.5 liters of 0.25 M Tris buffer, pH 7.5.
Useful information:
pKa Tris = 8.1
Formula we...