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?
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To calculate the pH of a solution, we need to consider the components of the solution and their interaction with water. In this case, we have tris(hydroxymethyl)aminomethane ("tris base") and tris hydrochloride dissolved in water.
First, let's calculate the final concentration of tris base and tris hydrochloride in the solution.
The molar mass of tris base is 121.14 g/mol. So, the number of moles of tris base is:
moles of tris base = mass of tris base / molar mass of tris base
= 10.0 g / 121.14 g/mol
= 0.0826 mol
The molar mass of tris hydrochloride is 157.60 g/mol. So, the number of moles of tris hydrochloride is:
moles of tris hydrochloride = mass of tris hydrochloride / molar mass of tris hydrochloride
= 10.0 g / 157.60 g/mol
= 0.0635 mol
Now, let's calculate the total concentration of tris base and tris hydrochloride in the solution. Since the volume of the solution is given as 0.250 L, we can use the formula:
concentration (molarity) = moles / volume
concentration of tris base = moles of tris base / volume of solution
= 0.0826 mol / 0.250 L
= 0.330 M
concentration of tris hydrochloride = moles of tris hydrochloride / volume of solution
= 0.0635 mol / 0.250 L
= 0.254 M
Next, let's calculate the moles of NaOH added to the solution. We know the volume and concentration of NaOH:
moles of NaOH = volume of NaOH (L) * concentration of NaOH (M)
= 0.0105 L * 0.500 M
= 0.00525 mol
Now, let's consider the reaction that occurs between NaOH and tris base. NaOH is a strong base and tris base is a weak base, so the NaOH will react with the tris base to form tris hydroxide.
NaOH + tris base → tris hydroxide
Since NaOH and tris base react in a 1:1 stoichiometric ratio, the moles of tris hydroxide formed will also be 0.00525 mol.
The final concentration of tris hydroxide will be the sum of the initial concentration of tris hydroxide and the moles formed in the reaction:
final concentration of tris hydroxide = initial concentration of tris hydroxide + moles of tris hydroxide formed
Since the initial concentration of tris hydrochloride is 0.254 M and the moles of tris hydroxide formed is 0.00525 mol, we can calculate the final concentration:
final concentration of tris hydroxide = 0.254 M + 0.00525 mol / 0.250 L
= 0.274 M
To calculate the pH of the solution, we need to consider that tris hydroxide is a weak base and will undergo partial ionization in water:
tris hydroxide ⇌ tris base + hydroxide ion
Since tris hydroxide is a base, its ionization constant is given by:
Kb = [tris base][hydroxide ion] / [tris hydroxide]
From the equation, we can see that the [tris base] and [tris hydroxide] will be equal. Thus, let's assume x be the concentration of tris base and tris hydroxide formed. Therefore, we can write:
Kb = x * x / x
= x
Now, we can use the Kb value of tris hydroxide to calculate x and determine the concentration of hydroxide ions.
Kb = [concentration of hydroxide ion]² / [concentration of tris hydroxide]
Rearranging the equation, we have:
[concentration of hydroxide ion]² = Kb * [concentration of tris hydroxide]
[concentration of hydroxide ion] = √(Kb * [concentration of tris hydroxide])
= √(4.0 * 10^-6 * 0.274)
= 8.3 * 10^-4 M
To find the pOH (the negative logarithm of the hydroxide ion concentration), we can use the formula:
pOH = -log([concentration of hydroxide ion])
= -log(8.3 * 10^-4)
= 3.08
Finally, to find the pH, we can use the equation:
pH = 14 - pOH
= 14 - 3.08
= 10.92
Therefore, the pH of the solution after the addition of 10.5 mL of 0.500 M NaOH will be approximately 10.92.
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).