Question

If the K_b of aniline is 4.3 x 10^-10, calculate the pH when 10.5 g of aniline is dissolved in 120 mL of water.

Select one:

a. 9.3

b. 12.8

c. 14.0

d. 2.9

Answer 1

OPTION "A" IS CORRECT

Aniline is an amine (containing NH2 on a benzene ring).

We have Kb(dissociation constant of a base) =4.3×10^–10

Let x be the degree if ionisation and c be the concentration (moles/volume).

Writing reaction we get,

At equilibrium

PhNH2 + H2O >< PhNH3+ + OH–

Writing their values which is present at equilibrium we have

PhNH2= c(1-x) , PhNH3+= cx and OH–= cx

So writing

Kb= (cx) (cx) /c(1–x)

But x is very small (as Aniline is a weak base) so 1–x~1

So we have

Kb~cx^2 or we can have finally

x=√Kb/c

We have to find c so we have to find no. Of moles of Aniline in 120 ml of water which on division will give c finally.

So molar mass of Aniline (PhNH2) = 93 g/mol

So no. Of moles of Aniline = 10.5/93=0.11 moles

So c= 0.11×1000/120 (since volume is given in ml and 1ml=1/1000 l).

c= 0.91mol/lit.

So x=√4.3×10^–10/0.91 = 10^–5 √4.3/0.91 = 2.2×10^–5

So [OH–]= cx = 0.91 × 2.2 × 10^–5 = 2.002 × 10^–5 ~2×10^–5 M

Also we know [H+] [OH–] = 10^–14

So, [H+]= 10^–14/[OH–] = 10^–14/2×10^–5 = 5 ×10^–10 M

So pH(of aniline) = –log[H+] = –log[ 5×10^–10](log at the base10) = –log(5) + 10= 10–0.69=9.31~ 9.3

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