What stoichiometric concentration of the indicated substance is required to obtain an aqueous solution with the pH} value shown below?
aniline (C6H5NH2) for PH of 8.90
C6H5NH2 HOH ==> C6H4NH3^+ + OH^-
Set up an ICE chart, substitute into Kb for aniline of
Kb = (C6H5NH2^+)(OH^-)/(C6H5NH2)
Convert pH to pOH, then to (OH^-) which equals (C6H5NH3^+) and solve for (C6H5NH2).
Post your work if you gets stuck.
0.147
To determine the stoichiometric concentration of aniline (C₆H₅NH₂) required to obtain an aqueous solution with a pH of 8.90, you need to consider the dissociation of aniline in water.
Aniline is a weak base, and when it dissolves in water, it accepts a proton (H⁺) to form the anilinium ion (C₆H₅NH₃⁺) and hydroxide ion (OH⁻) according to the following equilibrium reaction:
C₆H₅NH₂ + H₂O ⇌ C₆H₅NH₃⁺ + OH⁻
The pH of a solution is defined as the negative logarithm of the concentration of hydrogen ions (H⁺). In this case, we need to calculate the concentration of hydroxide ions (OH⁻) because aniline is a base.
To determine the concentration of hydroxide ions, we can use the ion product of water (Kw = [H⁺][OH⁻]). At 25°C, Kw is approximately 1.0 x 10^(-14) mol^2/L^2.
Since we are given the pH of 8.90, we can find the concentration of hydroxide ions as follows:
pOH = 14 - pH
pOH = 14 - 8.90
pOH ≈ 5.10
Now, we can calculate the concentration of hydroxide ions using the relation:
[OH⁻] = 10^(-pOH)
[OH⁻] = 10^(-5.10)
[OH⁻] ≈ 7.94 x 10^(-6) mol/L
Since aniline is a weak base, it does not dissociate completely, so we need to consider the stoichiometry of the reaction. In this case, the stoichiometric coefficient of aniline is 1.
Therefore, the stoichiometric concentration of aniline required to obtain an aqueous solution with a pH of 8.90 is approximately equal to the concentration of hydroxide ions:
[C₆H₅NH₂] ≈ 7.94 x 10^(-6) mol/L.