Question

What is the ph of a 0.25 m solution of c6h5nh2 given that its kb is 1.8 x 10-6?

Answer 1

The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.

How do we calculate pH of weak base?

pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:

pH = pKb + log([HB⁺]/[B])

pKb = -log(1.8×10⁻⁶) = 5.7

Chemical reaction for C₆H₅NH₂ is:

                          C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻

Initial:                     0.25                           0            0

Change:                    -x                             x             x

Equilibrium:        0.25-x                           x             x

Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]

Kb = x² / 0.25 - x

x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:

1.8×10⁻⁶ = x² / 0.25

x² = (1.8×10⁻⁶)(0.25)

x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]

On putting all these values on the above equation of pH, we get

pH = 5.7 + log(0.67×10⁻³/0.25)

pH = 3.13

Hence pH of the solution is 3.13.

To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361

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