Consider a solution that contains both C6H5NH2 and C6H5NH3+. Calculate the ratio [C6H5NH2]/[C6H5NH3+] if the solution has the following pH values. (Assume that the solution is at 25°C.)
(a) pH = 4.70
(b) pH = 5.26
(c) pH = 5.42
(d) pH = 4.96
To calculate the ratio [C6H5NH2]/[C6H5NH3+], we can use the Henderson-Hasselbalch equation, which tells us the relationship between the pH of a solution and its acid-base equilibrium. The Henderson-Hasselbalch equation is given as follows:
pH = pKa + log([A-]/[HA])
Where pH is the measured pH of the solution, pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, the weak acid is C6H5NH3+ (phenylammonium ion) and the conjugate base is C6H5NH2 (aniline).
The pKa value for the C6H5NH3+/C6H5NH2 pair is 4.71 at 25°C. So, by substituting the given pH values into the Henderson-Hasselbalch equation, we can determine the ratio [C6H5NH2]/[C6H5NH3+] for each case.
(a) pH = 4.70:
We can use the Henderson-Hasselbalch equation:
4.70 = 4.71 + log([C6H5NH2]/[C6H5NH3+])
log([C6H5NH2]/[C6H5NH3+]) = 4.70 - 4.71 = -0.01
Taking the antilog of both sides:
([C6H5NH2]/[C6H5NH3+]) = 10^(-0.01) ≈ 0.794
(b) pH = 5.26:
Using the Henderson-Hasselbalch equation:
5.26 = 4.71 + log([C6H5NH2]/[C6H5NH3+])
log([C6H5NH2]/[C6H5NH3+]) = 5.26 - 4.71 = 0.55
Taking the antilog of both sides:
([C6H5NH2]/[C6H5NH3+]) = 10^(0.55) ≈ 3.56
(c) pH = 5.42:
Using the Henderson-Hasselbalch equation:
5.42 = 4.71 + log([C6H5NH2]/[C6H5NH3+])
log([C6H5NH2]/[C6H5NH3+]) = 5.42 - 4.71 = 0.71
Taking the antilog of both sides:
([C6H5NH2]/[C6H5NH3+]) = 10^(0.71) ≈ 5.01
(d) pH = 4.96:
Using the Henderson-Hasselbalch equation:
4.96 = 4.71 + log([C6H5NH2]/[C6H5NH3+])
log([C6H5NH2]/[C6H5NH3+]) = 4.96 - 4.71 = 0.25
Taking the antilog of both sides:
([C6H5NH2]/[C6H5NH3+]) = 10^(0.25) ≈ 1.78
So, the ratios [C6H5NH2]/[C6H5NH3+] for the given pH values are approximately:
(a) 0.794
(b) 3.56
(c) 5.01
(d) 1.78
To determine the ratio [C6H5NH2]/[C6H5NH3+], we need to consider the acid-base equilibrium of C6H5NH2 and C6H5NH3+ in water. The reaction is as follows:
C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-
From this, we can write the expression for the equilibrium constant (Kw):
Kw = [C6H5NH3+][OH-] / [C6H5NH2]
At 25°C, the value of Kw is 1.0x10^-14. Since OH- concentration in water is usually very small, we can assume it to be negligible. Therefore, the equation can be simplified to:
Kw ≈ [C6H5NH3+][OH-] / [C6H5NH2] ≈ [C6H5NH3+]/[C6H5NH2]
Taking the negative logarithm of both sides of the equation gives:
-log(Kw) ≈ -log([C6H5NH3+]/[C6H5NH2])
Using the definition of pH (pH = -log[H+]), we can write the equation as:
14 ≈ -log([C6H5NH3+]/[C6H5NH2])
Simplifying further, we get:
log([C6H5NH3+]/[C6H5NH2]) ≈ -14
Now we can calculate the ratio [C6H5NH2]/[C6H5NH3+] using the given pH values.
(a) pH = 4.70:
14 ≈ -log([C6H5NH3+]/[C6H5NH2])
14 ≈ -log([C6H5NH3+]/[C6H5NH2])
log([C6H5NH3+]/[C6H5NH2]) ≈ -14
[C6H5NH3+]/[C6H5NH2] ≈ 10^-14
(b) pH = 5.26:
log([C6H5NH3+]/[C6H5NH2]) ≈ -14
[C6H5NH3+]/[C6H5NH2] ≈ 10^-14
(c) pH = 5.42:
log([C6H5NH3+]/[C6H5NH2]) ≈ -14
[C6H5NH3+]/[C6H5NH2] ≈ 10^-14
(d) pH = 4.96:
log([C6H5NH3+]/[C6H5NH2]) ≈ -14
[C6H5NH3+]/[C6H5NH2] ≈ 10^-14
In all the given pH values, the ratio [C6H5NH2]/[C6H5NH3+] is approximately equal to 10^-14.
pKb = 8.77
use the Handerson-Hasselbalch equation
pOH = pKb + log [C5H5NH+]/ [C5H5N]
1
pOH = 14 - 4.50= 9.50
9.50 - 8.77=0.73
10^0.73 = 5.4 = [C5H5NH+]/ [C5H5N]
[C5H5N]/ [C5H5NH+]= 1 / 5.4=0.19