Respuesta :
Answer:
a) 0.0323 M
b) ΔpH = 0.10
Explanation:
A buffer is a solution of a weak acid and its conjugate base, or of a weak base and its conjugate acid. This solution keeps the pH almost constant because it has equilibrium between the acid and the base. So, when acid is added, it reacts with the base and when a base is added, it reacts with the acid.
To calculate the pH in a buffered solution, we may use the Henderson-Hasselbalch equation:
[tex]pH = pKa + log\frac{[A^-]}{[HA]}[/tex]
Where pKa is the -logKa and Ka is the acid constant, and [A⁻] is the concentration of the base (or ionic compound of the acid), and [HA] is the concentration of the acid.
a) C₆H₅NH₂ is the base and its conjugate acid is C₆H₅NH₃⁺. pKa + pKb = pKw, where pKw is from the water equilibrium and is equal to 14.
pKa = 14 - 9.13 = 4.87
So,
pH = pKa + log [C₆H₅NH₂]/[C₆H₅NH₃⁺]
5.75 = 4.87 + log (0.245)/[C₆H₅NH₃⁺]
5.75 - 4.87 = log (0.245) - log [C₆H₅NH₃⁺]
log[C₆H₅NH₃⁺] = -0.6108 - 0.88
log [C₆H₅NH₃⁺] = -1.4908
[C₆H₅NH₃⁺] = [tex]10^{-1.4908}[/tex]
[C₆H₅NH₃⁺] = 0.0323 M
b) First, let's calculated the concentration of NaOH, knowing that the molar masses are
MNa = 23 g/mol
MO = 16 g/mol
MH = 1 g/mol
MNaOH = 40 g/mol
n = m/M
n = 0.360/40 = 0.009 mol
The concentration is: n/V = 0.009/1.5 = 0.006 M.
The base reacts with the acid to form more base compound. Than the concnetration of aniline and aniline hydrochloride will be know:
[C₆H₅NH₂] = 0.245 + 0.006 = 0.2510 M
[C₆H₅NH₃⁺] = 0.0323 - 0.006 = 0.0263 M
So, the new pH will be:
pH = 4.87 + log(0.2510)/(0.0263)
pH = 4.87 + 0.98
pH = 5.85
ΔpH = 5.85 - 5.75
ΔpH = 0.10
Concentration of C₆H₅NH₃⁺ in the solution is 0.0323 M and change in pH of the buffer solution due to NaOH is 0.10.
What is buffer solution?
A buffer is a solution of a weak acid and its conjugate base, or of a weak base and its conjugate acid, it maintain the pH of the solution.
(1). Concentration of C₆H₅NH₃⁺ is calculated as follow:
From the question it is clear that C₆H₅NH₂ is a base and C₆H₅NH₃⁺ is its conjugate acid, we also know that pKw = pKa + pKb, where
pKw = 14 (from water equilibrium)
pKb = 9.13 (given)
pKa = 14 - 9.13 = 4.87
We will use Henderson - Hasselbalch equation to calculate the concentration of C₆H₅NH₃⁺:
[tex]${\rm{pH = pKa + log}}\frac{{\left[ {{{\rm{A}}^{\rm{ - }}}} \right]}}{{\left[ {{\rm{HA}}} \right]}}$[/tex]
[A⁻] = concentration of base = 0.245 M (given)
[HA] = concentration of acid = to find?
pH = 5.75 (given)
Putting all the values in the pH equation we get:
5.75 = 4.87 + log(0.245) - log[C₆H₅NH₃⁺]
log[C₆H₅NH₃⁺] = -0.6108 - 0.88
log[C₆H₅NH₃⁺] = -1.4908
Or [C₆H₅NH₃⁺] = [tex]${10^{ - 1.4908}}$[/tex]
[C₆H₅NH₃⁺] = 0.0323 M
(2). To calculate the change in pH of the solution first we calculate the concentration of added NaOH:
C = n / V
Where, n = no. of moles and V = volume = 1.5 L (given)
n = W/ M
where, W = 0.360 g and M = 40 g/mol
n = 0.360/40 = 0.009 mol
Therefore, C = 0.009/1.5
Concentration of C₆H₅NH₂ = 0.245 + 0.006 = 0.2510 M
Concentration of C₆H₅NH₃⁺ = 0.0323 - 0.006 = 0.0263 M
New pH will be calculated using Henderson - Hasselbalch equation:
pH = 4.87 + log(0.2510)/(0.0263)
pH = 5.85
Therefore change in pH = 5.85 - 5.75 = 0.10
Hence, 0.0323 M is the Concentration of C₆H₅NH₃⁺ in the solution and change in pH is equal to 0.10.
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