Question

the base-dissociation constant for aniline, c6h5nh2, is 7.5 ✕ 10−10, at 25°c. calculate the ph of a 0.490 m aniline solution in water at 25°c.

Answer 1

To find the pH of a 0.490 M aniline solution, we use an ICE table and the given base-dissociation constant. We calculate the concentration of OH- produced at equilibrium, from which we calculate pOH. Finally, we use the relationship between pOH and pH to determine the pH of the solution.

Step-by-step explanation:

To calculate the pH of a 0.490 M aniline solution, we need to use the given base-dissociation constant (Kb) for aniline,

Kb = 7.5 × 10−10.

The chemical equation for aniline accepting a proton from water is:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH−.

Using an ICE table, the initial concentration (in molarity) of C6H5NH2 is 0.490 and the initial concentrations of C6H5NH3+ and OH− are both 0. For the equilibrium, we will have:

C6H5NH2 = 0.490 - x

C6H5NH3+ = x

OH− = x

Since Kb is small, x will be much smaller than 0.490, so we can approximate 0.490 - x as 0.490. Kb is then:

Kb = [C6H5NH3+][OH−] / [C6H5NH2]

7.5 × 10−10 = (x)(x) / (0.490)

We solve for x, which gives us the concentration of OH−. From there, we can calculate pOH = -log[OH−] and use the relation pH + pOH = 14 to find the pH.